ELECTRIC FIELD Example: Solution: A ing-shaped conducto with adius a caies a total positive chage Q unifomly distibuted on it. Find the electic field at a point P that lies on the axis of the ing at a distance x fom its cente. The figue shows the electic field de at the point P due to an infinitesimal segment at the top of the ing. The x-and y-components of de is de cos Ө and de sin Ө. If we take anothe segment of the bottom, we see that its contibution to the field at P has the same x-component but opposite y-component. Hence total y-component becomes zeo. Hence the field at P will be along x-axis. Theefoe, the net field at P is E decos 1 dq x.. 4 x a x a 4 x a x 3 4 x a o, E Qx 3 4 x a Qx dq 3 (1) At the cente of the ing, x, E i
1 Q () If x >> a, E i 4 x This field is same as that of a point chage at the cente. (3) E will be maximum at the points whee de dx o, x a, and E max 1 Q. 4 3 3a The gaph of the vaiation of E with x is as shown in figue. The negative value of E indicates that the diection of the field is opposite to the positive value. (4) If a >> x, E Q 4 a a 3 x, x whee is the linea chage density. The foce on a negative point chage q is F kx, q x a q whee k a
On eleasing the point chage ( q) of mass m, it will execute SHM (if x << a) with the time-peiod. T m k a m q It should be noted that if x is not negligible elative to a, the negative chaged paticle will oscillate about the cente but the motion will not be simple hamonic. Calculations of electic field due to continuous chage distibution. (1) Linea chage distibution Infinite line of chage: x d tan Ө dx d sec Ө dө de P 1 dq 4 1 dx 4 x d 1 dsec d de P 4 d tan d
1 de P d 4 d E decos dcos 4 d 4 d sin / / sin sin 4 d E d Semi infinite wie E 1 4 d E 11 4 d E net 4 d E 1 E 11 4 d
E net 4 d Fo finite wie sin sin 4 d E 1 cos cos 4 d E 11 Electic field fo unifomly chaged ac dl R dө dq l dl mrdө de 1 dq 4 R 1 Rd 4 R de d 4 R E decos
4 R / / cos d E sin R ф in adians. ELECTROSTATIC POTENTIAL Electostatic potential is the scala way of epesenting the egion aound any electic chage configuation (as the electic field was the vecto way of epesentation) which is useful in calculating the wok involved when a chage moves aound othe chages o field. When a chage is moved slowly aound anothe chage, some electical wok may be done and the wok done by the electic field is equal and opposite to the wok done by extenal agent duing the same movement. Physically you know that high potential means high capacity to do wok. Close you ae to a positive chage, highe is the potential expeienced by you (means the positive chage will do lage wok in pushing a unit chage fom that point to infinity). Close you ae to a negative chage, lesse is the potential expeienced by you. SI unit fo electic potential that follows fom its definition is the "joule pe coulomb". This combination is given a special name, called "volt" in honou of the Italian scientist Alessando Volta (1745-187). 1 volt 1 V 1 J/C 1 joule/coulomb
(i) (ii) (iii) (iv) (v) Test chage is an infinitesimal positive chage. Potential is a scala quantity. Potential may be positive, negative o zeo. The potential at a point in the electic field due to an isolated positive chage will be positive and that due to an isolated negative chage will be negative elative to zeo potential at infinity. Potential diffeences ae fundamental since they do not depend on defining an abitay efeence point and assigning an abitay potential enegy to it. But the potential at a point depends on defining an abitay efeence point and assigning an abitay potential enegy to it. (vi) The potential diffeence V a V b is called potential of a with espect to b. Sometimes it is abbeviated as V ab i.e., V ab V a V b. (vii) (viii) When a chage q moves fom i to f in an electic field, the wod done by the electic field on the chage q is W ele q(v i V f ). Fo a positive chage, W > if V i > V f. Hence a positive chage moves fom high potential to low potential if it is fee to move. Fo a negative chage, W > if V i < V f. Hence a negative chage moves fom low potential to high potential if it is fee to move. The wok done by the extenal foce is W ext q(v f V i ), (note the ode of subscipts), if the chage q moves fom i to f in a electic field vey slowly. Potential due to a Point Chage It is the wok done by an extenal agent in moving unit positive chage slowly fom infinity to the point whee potential is to be calculated. Figue shows an isolated positive point chage of magnitude q. We want to find potential V due to q at a point P, at a adial distance fom that chage.
This potential will be detemined elative to zeo potential at infinity. Let us suppose that a test chage (positive) q moves fom infinity to point P. Since the wok done by the electic field is independent of the path followed by the test chage, we take the simplest path along the adial line fom infinity to q passing though P. Conside an abitay position P' at a distance ' fom q. The electostatic foce on q at P' is 1 qq F, diected adially away. Fo a futhe diffeential 4 ' displacement ds, the wok done by the foce F is dw F.ds Since the diection of ds is opposite to the diection of inceasing '. Hence ds d ' So, dw Fdc 1 qq 4 ' d ' o, W 1 d ' qq 4 ' W 1 1 qq 4 ' 1 1 1 qq 4 qq 4
W q V q 4 [potential due to a point chage] If the point chage is negative the electostatic foce points towads q, and hence W will be positive, giving a negative potential V. Thus the sign of V is same as the sign of q, when we assign zeo potential at infinity. Potential due to an Electic Dipole Figue shows a dipole of dipole moment p qd. We have to find the electic potential at the point P, distant fom the cente O of the dipole. The line OP makes an angle Ө with the dipole axis. The potential at P due to the positive chage (+ q) is V 1 q 4 1 and that due to the negative chage ( q) is V 1 q 4 The net potential at P is V V 1 + V 1 q q 4 1 q 1 (i) 4 1
Example: Solution: Fo a small dipole, we have usually, >> d, whee "d" is the sepaation between the two chages. Unde these conditions, We may wite, 1 - d cos Ө and 1 - using these quantities in equation (i), we get V q dcos. 4 Infinite numbe of same chage q ae placed at x 1,, 4, 8... What is the potential at x? V 1 q q q q... 4 1 4 8 q 1 4 1 1 Example: Solution: q 4 q If the altenative chages ae unlike, then what will be the potential? Then, 1 q q q q V... 4 1 4 8
1 q q q q...... 4 1 8 1 1 1 1 4 1 1 1 1 4 4 1 q 4 3 Potential diffeence: Example: The wok done in taking a chage fom one point to the othe in an electic field is called the potential diffeence between two points. Thus, if w be wok done in moving a chage q fom B to A then the potential diffeence is given by V A V B (i) (ii) (iii) (iv) W q Unit of potential diffeence is volt. This is a scala quantity. Potential diffeence does not depend upon Co-odinate system. Potential diffeence does not depend upon the path followed. This is, because electic field is a consevative foce field and wok done is consevative foce field does not depend upon path followed. In the following fig. Along which path the wok done will be maximum in caying a chage fom A to B in the pesence of any anothe chage.
Solution: Same fo all the path [Because the wok done doesn't depend upon the path] Example: Solution: A chage µc is situated at the oigin of X-Y plane. What will be potential diffeence between points (5a, ) and ( 3a, 4a) Distance between (, ) & (5a, a), 1 5a V 1 5a kq 5a Distance between (, ) & ( 3a, 4a) 5a 9a V kq 5a V 1 V 16a Relationship between electic potential and intensity of electic field (i) V A A E.d,
(ii) V A electic potential at point A. Potential diffeence between two points in an electic field is given by negative value of line integal of electic field i.e., B V B V A E.d A (iii) E V gad (gadient) (iv) i j k x y z E x E y E z V, x V, y V z If v is a function of only, then E dv d (v) Fo a unifom electic field, E decease in the value of V. V and it's diection is along the Example: Electic potential fo a point (x, y, z) is given by V 4x volt. Electic field at point (1,, ) is
Solution: Example: Solution: E dv dx 8x E at (1,, ) 8 V/m Magnitude of E 8V/m diection along x axis. 1 Electic field is given by E x and x m will be E dv dx dv Edx potential diffeence between x 1 B A dv B A E.dx 1 V V x B A 1 5 volts Potential diffeence 5 volt.
Example: Solution: The potential at a point (x,, ) is given as V 1 15 5 3. What will be electic field intensity at x x x x 1m? E V V V V i j k x y z o ie x + je y + ke z V V V i j k x y z V x V y V z Compaing both sides E x V x 1 15 5 x x x x 3 1 15 3 5 3 4 x x x Fo x 1, (E x ) 55 V/m
Example: Solution: What will be the electic field intensity at 3. Fo < < 4, V 5 volts dv E d In the above poblem, what will value of E at 6? at 7m V volt at 1 5m V 1 4 volt E V V 1 1 4 7 5 1 volt/mete. Example: An oil dop 'B' has chage 1.6 x 1 19 C and mass 1.6 1 14 kg. If the dop is in equilibium position, then what will be the potential diff. between the plates. [The distance between the plates is 1mm]
Solution: Fo equilibium, electic foce weight of dop qe mg V q. mg d V mgd q 1.6 1 9.8 1 1 19 1.6 1 V 1 4 volt 14 3 [When a chaged paticle is in equilibium in electic field, the following fomula is often used qe mg] Equipotential Suface (i) (ii) (iii) (iv) (v) (iv) These ae the imaginay suface (dawn in an electic field) whee the potential at any point on the suface has the same value. No two equipotential sufaces eve intesects. Equipotential sufaces ae pependicula to the electic field lines. Wok done in moving a chage fom a one point to the othe on an equipotential suface is zeo iespective of the path followed and hence thee is no change in kinetic enegy of the chage. Component of electic field paallel to equipotential suface is zeo. Neae the equipotential sufaces, stonge the electic field intensity.
Example: Some equipotential sufaces ae shown in fig. What is the coect ode of electic field intensity? Solution: E B > E C > E A, because potential gadient at B is maximum. POTENTIAL ENERGY OF CHARGED PARTICLE IN ELECTRIC FIELD (i) (ii) (iii) Wok done in binging a chage fom infinity to a point against the electic field is equal to the potential enegy of that chage. Potential enegy of a chage of a point is equal to the poduct of magnitude of chage and electic potential at that point i.e. P.E. qv. Wok done in moving a chage fom one point to othe in an electic field is equal to change in it's potential enegy i.e. wok done in moving Q fom A to B qv B qv A NOTE: (iv) U B U A Wok done in moving a unit chage fom one point to othe is equal to potential diffeence between two points. Cicumfeence of the cicle in above example can be consideed as equipotential suface and hence wok done will be zeo. Potential Enegy of System:
(i) (ii) (iii) (iv) The electic potential enegy of a system of chages is the wok that has been done in binging those chages fom infinity to nea each othe to fom the system. If a system is given negative of it's potential enegy, then all chages will move to infinity. This negative value of total enegy is called the binding enegy. Enegy of a system of two chages 1 qq 1 PE 4 d Enegy of a system of thee chages PE 1 qq q q q q 4 1 3 3 1 1 3 31 (v) Enegy of a system of n chages. PE n n 1 1 qj. q i 4 i1 j1 ij i j Electic Field Example: Solution: Two symmetical ings of adius R each ae placed coaxially at a distance R mete. These ings ae given the chages Q 1 & Q espectively, unifomly. What will be the wok done in moving a chage q fom cente of one ing to cente of the othe. i Wok done q (V V 1 ) potential at the cente of fist ing
1 Q1 1 Q V 1 4 R 4 R R 1 Q Q 1 4R potential at the cente of second ing 1 Q 1 Q1 V 4R R 4 R R 1 Q Q 1 4R wok done q (V V 1 ) q Q Q1 Q1 Q 4 R NOTE: q 1 W Q Q 1 4 R 1 1 Wok done in moving the same chage fom second to the fist ing will be negative of the wok done calculated above i.e. q 1 W Q Q 1 4 R 1 1 ELECTRIC DIPOLE (i) A system consisting of two equal and opposite chages sepaated by a small distance is temed an electic dipole.
(ii) (iii) Example: Na + Cl, H + Cl etc. An isolated atom is not a dipole because cente of positive chage coincides with cente of negative chage. But if atom is placed in an electic field, then the positive and negative centes ae displaced elative to each othe and atom become a dipole. DIPOLE MOMENT: The poduct of the magnitude of chages and distance between them is called the dipole moment. (a) (b) (c) This is a vecto quantity which is diected fom negative to positive chage. Unit: Coulomb mete (C-M) Dimension: [M L 1 T 1 A 1 ] (d) It is denoted by p that is p qd Electic field due to a dipole (i) Thee ae two components of electic field at any point (a) E in the diection of (b) E in the diection pependicula to 1 P cos E. 3 4
(ii) Resultant q P sin E. 3 4 E E E P 4 3 1 3cos (iii) Angle between the esultant E and, is given by E Ө tan 1 E 1 tan 1 tan (iv) If Ө, i.e. point is on the axis (v) 1 P E axis. 3 4 Ө, i.e. along the axis. If Ө 9, i.e. point is on the line bisecting the dipole pependiculaly 1 P E equato. 3 4 (vi) So, E axis E equato (fo same ) (vii) E axis 1 P. 4
(viii) V 1 P E equato. 4 whee P q. 3 / 1 q cos. 4 4 1 P cos. 1 P.. 4 1 P.. 3 4 whee Ө is the angle between P and. NOTE : (ix) If Ө 9, V equato (x) Hee we see that V but E fo points at equato. (i) (ii) This is not essential that at a point, whee E, V will also be zeo thee eg. inside a unifomly chaged sphee, E but V Also if V, it is not essential fo E to be zeo eg. in equatoial position of dipole V, but E Electic Dipole In an Electic Field Unifom Electic Field (i) When an electic dipole is placed in an unifom electic field, A toque acts on it which subjects the dipole to otatoy motion. This Ʈ is given by Ʈ PE sin Ө
o P E (ii) Potential enegy of the dipole U PE cosө P.E 1 p cos o, V. 4 *potential due to an electic dipole+ whee p qd is the magnitude of the electic dipole moment. Potential due to a System of Chages Suppose a system contains chages q 1, q, q 3... q n, at distances 1,, 3... n, fom a point P. The potential at P, 1 q1 Due to q 1 is V 1 4 1 1 q Due to q is V 4 Due to q 3 is V 3 and so on. 1 q 4 3 3 By the supeposition pinciple, the potential V at P due to all the chages is the algebaic sum of potentials due to the individual chages. Theefoe, net potential at P is V V 1 + V + V 3 + + V n
1 q1 q q3 q n o, V... 4 1 3 n Example : In the given figue, thee ae fou point chages placed at the vetices of a squae of side a 1.4 m. If q 1 + 18 nc, q 4 nc, q 3 + 35 nc and q 4 + 16 nc, then find the electic potential at the cente P of the squae. Assume the potential to be zeo at infinity. Solution: The distance of the point P fom each chage is a 1.4m 1.4 1m V V 1 + V + V 3 + V 4 1 q q q q 4 1 3 4 1 3 4 1 q q q q. 4 1 3 4 9 1 18 4 35 16 1 C. 4 1m (9 1 9 ) (45 1 9 ) V 45 V
Example: Solution: Two point chages 5 µc and + 3 µc ae placed 64 cm apat. At what points on the line joining the two chages is the electic potential zeo. Assume the potential at infinity to be zeo. Fo the potential to be zeo, at the point P, it will lie eithe between the two given point chages o on the extended position towads the chage of smalle magnitude. Suppose the point P is at a distance x cm fom the chage of lage magnitude. Then fom figue (a) 6 6 1 5 1 3 1. 4 x 1 64 x 1 o, o, 3 5 64 x x 8x 3 x 64 x o, x 4 cm Fom the figue (b), we get 3 5 x 64 x
3 x o, x x 64 o, x 16 cm Potential due to a Continuous Chage Distibution When a chage distibution is continuous, instead of consisting of sepaate point chages, we divide it into small elements, each caying a chage dq. Consideing this diffeential element of chage dq as a point chage, we detemine the potential dv at the given point P due to dq, and then integate ove the continuous chage distibution. Assuming the zeo of potential to be at infinity, dv 1 dq 4 Whee is the distance of P fom dq. The potential V at P due to the entie distibution of chage is given by V dv 1 dq 4 Case I : Line chage distibution Fo the chage unifomly distibuted along a line dq dl, whee is the linea chage density and dl is a line element. Case II : Suface chage distibution Fo the chage unifomly distibuted on a suface with suface chage density, the chage on a suface element da is dq da Case III: Volume chage distibution Fo the chage unifomly distibuted in a volume with volume chage density, the chage in a volume element dv is dq dv.
Example : Electic chage Q is unifomly distibuted aound a thin ing of adius a. Find the potential at a point P on the axis of the ing at a distance x fom the cente of the ing. Solution: NOTE: Since the entie chage is at distance x a fom the point P, theefoe, the potential at P due to the total chage Q is V 1 Q 4 x a At the cente, V 1 Q 4 a Vaious Cases: A thin unifomly chaged spheical shell (a) Potential at an outside point: We aleady know that fo all points outside the sphee, the field is calculated as if the sphee wee emoved and the total chage q is positioned as a point chage at the cente of the sphee.
Remembe: Hence, the potential at an outside point at a distance fom its cente is same as the potential due to a point chage q at the cente. i.e., V 1 q 4 *fo an outside point+ This is valid only when the chage distibution is unifom. (b) Potential at an inside point: We know that inside the spheical shell of unifom distibution of chage, the electic field E is zeo eveywhee. Hence, if a test chage q is moved inside the sphee no wok is done on that chage against the electic foce. Hence fo any infinitesimal displacement anywhee inside the sphee, W o q V o, V. Hence V (potential) emains constant inside the sphee and is equal to its value at the suface i.e., V 1 q 4 R [fo an inside point, < R] A Chaged Conducting Sphee Since the chage on a conducting sphee is unifomly distibuted on its suface, theefoe, it acts just like a spheical shell of chage unifomly distibuted in it. So V 1 q 4
...[fo an outside point, > R] and V 1 q 4 R...[fo an inside point, < R] Impotant Points about Conducting Sphee: Fo a chaged spheical shell (1) V inside V suface RE suface constant () V outside E outside It inceases o deceases with inceasing depending on whethe q < o q >. (3) Figue shows the vaiation of V with fo a positively chaged spheical shell. V is maximum inside and on the suface of the sphee. (4) Figue shows the vaiation of V with fo a negatively chaged spheical shell. V is minimum inside and on the suface of the sphee. It is maximum ( ) at infinity. (5) The maximum potential (V m ) upto which a spheical shell can be aised is V m RE m whee E m is the electic field magnitude at which ai becomes conductive (known as dielectic stength of ai). Its value is about 3 1 6 N/C.
Thus, a conducting sphee of 1 m adius can be chaged upto a maximum potential of 3 1 6 volt in ai. Potential Due to Unifomly Distibuted Chage in a Spheical volume Case I: At a point P outside the sphee: As we know fo all extenal points, the total chage q is supposed to be concentated at the cente of the sphee. So the potential at an outside point P, distant fom the cente O is V P 1 q 4 potential at an outside point i.e., fo > R Case II: At a point P inside the sphee: Let us divide the entie sphee into two pats, one is a sphee of adius OP and the othe is the spheical shell of intenal adius and extenal adius R. As in case I, the potential at P due to the chage of sphee of adius is V 1 1 q' 4 4 3 1 3 4
3 whee q 4 R 3 3 Fo finding the potential due to the othe pat, let us conside a thin spheical element of adius x and thickness dx. Its volume is 4πx dx, and hence it contains the chage dq (4πx dx)p 4πpx dx The potential at P due to this element is dv 1 dq 4 x 1 4x dx 4 x 1 xdx The potential at P due to the spheical shell of intenal adius and extenal adius R is V dv xdx x R R
So, the net potential at P is V V 1 + V R 3 3R 6 q 3R 4 R 3 6 3 q o, V 3 3R 8 R.[potential at an intenal point i.e., < R] (1) Putting R, we get V suface q 4 R () Putting, we get V cente 3q 8 R 3 V suface (3) Figue shows the Vaiation of V with. Fo < < R, the gaph is paabolic.
Fo > R, the gaph is hypebolic. EQUIPOTENTIAL SURFACES "An equipotential suface is a suface on which the electic potential V is same at evey point". The electic field in a egion can be epesented by a family of equipotential sufaces, each of which coesponds to diffeent potential. Figue (a) and (b) show diffeent aangement of chages and coesponding field lines, and coss-sections of equipotential sufaces fo (a) the unifom electic field due to a lage plane sheet of chage (b) the electic field associated with a positive point chage. Figue (c) epesents the equipotential sufaces fo an electic dipole. Figue (d) epesents the equipotential sufaces fo two identical positive chages. Example : Some equipotential sufaces ae shown in the figue below. What can you say about the magnitude and the diection of the electic field? Solution: Fist we will find the diection of the E noting that the E lines ae always pependicula to the local equipotential sufaces and points in that diection in which the potential is deceasing. So it will be as shown below
Note that we have dawn the E lines pependicula to all equipotential sufaces and it is pointing in that diection in which the potential is deceasing. So, the total angle E is making with the positive x-axis is 1 as shown. Now, how to calculate the magnitude of the electic field! Fo this we need to jump fom one equipotential suface to anothe though shotest (pependicula) distance. Then calculate, E Change in potential Distance tavelled Let us suppose we ae jumping fom V line to 1 V line fom B to A. Note that the BA line is the shotest length between the two equipotential sufaces. The potential diffeence between the two lines is ( V 1 V) 1 volt. Now, calculate the length AB 1 sin 3 [cm] 5 cm 5 1 m Theefoe, 1 volt E 5 1 m [volt/m] Hence, the complete answe is that the E has the value V/m and makes 1 with positive x-axis. Example :
The electic potential existing in space is V(x, y, z) B(xy + yz + zx). Find the expession fo the electic field at point P(1, 1, 1) and its magnitude if B 1 S.I. unit. Solution: The fomula E dv d V V V E i j k x y z whee, V V V, and x y z in extended fom is witten as follows: ae patial deivatives of potential V with espect to x, y and z espectively. Fo patial deivative you should note that if say you ae V calculating, then othe vaiables y and z will be teated as constant. x Now, If we wite E iex jey ke z, then E x E y V, x V y V and E z. z In the question
V B(xy + yz + zx) So, E x V x B(y + z) because vaiables y and z ae teated constant. Now, put the values of B, y and z. Given B 1 and (x, y, z) ae (1, 1, 1) espectively. So, E x V x B(y + z) 1 (1 + 1) N/C Similaly, E y V y B(x + z) [now teating x and z constant] theefoe E y 1 (1 + 1) N/C and E z V z B (x + y)
N/C theefoe, E Ex i Ey j Ezk o, E i j k N / C Now, if we want the magnitude of E, then it's equal to E E E x y z 3 Example: An electic field E i j3 netwon/coulomb exists in the space. If the potential at the oigin is taken to be zeo, find the potential at (m, m). Solution: We have E dv d so, it can be witten in vecto fom as dv E.d Note, you can wite E as Ex i Ey j Ez k
and d dxi dy j dzk theefoe, E.d E x.dx E y.dy E z.dz In the given question the z component of E o the point is not given. So you can wite E.d E.dx E.dy Now, dv x E.d i 3 j. dxi dy j o, dv dx 3 dy y Now we will have to integate it within limits. Given V when x and y (lowe limit) and we have to calculate V when x and y (uppe limit) theefoe, V x y dv dx 3 dy x y V o, V x 3y o, V ( ) 3 ( ) o, V 4 6 1 volt. POTENTIAL ENERGY OF A SYSTEM OF CHARGES
"The electic potential enegy of a system of fixed point chages is equal to the wok that must be done by an extenal agent to assemble the system, binging each chage in fom an infinite distance". Let us fist conside a simple case of two point chages q 1 and q sepaated by a distance. To find the electic potential enegy of this two-chage system, we conside the chages q 1 and q initially at infinity. When one of the two chages (say q 1 ) is bought fom infinity to its place (say A), no wok is done, because no electostatic foce acts on it. But when we next bing the othe chage q fom infinity to its location (B), we must do wok because no, exets an electostatic foce on q duing its motion. Assuming the potential at infinity to be zeo, the potential at B due to the chage q 1 at A is 1 q1 V 4 The wok done by the extenal agent in binging the chage q fom infinity to B is W q V 1 qq 4 1 So, the potential enegy is U W 1 qq 4 1
1 qq 1 This fomula U 4 is tue fo any sign of q 1. If q 1 and q ae of same sign, q 1. q >, so the potential enegy U is positive. If q 1 and q ae of opposite signs, q 1 q <, and hence U is negative. Now, we conside a thee-chages system shown in the figue. Fist, we suppose that thee chages ae at infinity. Then we bing them one by one to thei locations. As befoe, the wok done by the extenal agent in binging the two chages q 1 and q (one by one is) 1 qq 1 W 1 4 1 (ii) Now, the potential at C, due to q 1 (at A) and q (at B) is V 1 q1 q 4 13 3 So, when q 3 is bought fom infinity to C, the wok done on it by the extenal agent is W q 3 V 1 q1q 3 qq 3 4 13 3 The total wok done is W W 1 + W 1 qq 1 q1q 3 qq 3 4 1 13 3 Hence, the potential enegy of the system is U W
1 qq 1 q1q 3 qq 3 4 1 13 3 o, U U 1 + U 13 + U 3 This can be genealized fo any system containing n point chages q 1, q, q 3...q n. The potential enegy of the system will be witten as U (U 1 + U 13 +. + U 1n ) + (U 3 + + U n ) o, U i j U ij + (U 34 + + U 3n ) +. + U (n 1)n 1 q i.q 4 i j ij j Thus, the potential enegy of a system of chages is the sum of potential enegies of ai possible pais of chages that can be fomed by taking chages fom the system. POTENTIAL ENERGY IN AN EXTERNAL FIELD (i) Potential Enegy of a single chage: The wok done by an extenal agent in binging a chage g fom infinity to the pesent location in the electic field is stoed in the fom of potential enegy of g. If V be the electic potential at the pesent location of the point chage g, then the wok done in binging it fom infinity to the pesent location is qv. The potential enegy of q at in an extenal field is U q..*potential enegy+
whee V is the potential at. (ii) Potential enegy of a system of two chages: Suppose two chages q 1 and q ae located at 1 and espectively in an extenal field. To find the potential enegy of the two chage system, fist bing q 1 fom infinity to. 1 The wok done in this pocess is Example : W 1 q 1 V( 1 ), whee V( 1 ) is the potential at 1 due to the extenal field E. Next, we bing q fom infinity to. The wok done on q against the extenal field E is W q V( ), whee V( ) is the potential at due to the extenal field E only. The wok done on q against the field of q 1 is qq 1 W 3, 4 1 whee 1 is the sepaation between q 1 and q. The potential enegy of the two-chage system is U W 1 + W + W 3 qv qq q V 4 1 1 1 Two potons ae sepaated by a distance Ft. What will be the speed of each poton when they each infinity unde thei mutual epulsion?
Solution: Let the chage on each poton be Q. Initially they ae at est, so thei kinetic enegy (KE) and they have only potential enegy equal to Q. 4 R Finally at infinity, they have infinite sepaation so thei potential enegy will become zeo. (because R now.) And total kinetic enegy is, ( 1 mv fo each poton) x 1 mv mv (whee m is the mass of each poton and v is the speed with which they ae moving at infinity). Now, total initial enegy total final enegy o, Q 4 R mv Theefoe, v Q. 4 mr Example: Solution: A bullet of mass gm is moving with a speed of 1 m/s. If the bullet has a chage of mico-coulomb, though what potential it be acceleated stating fom est, to acquie the same speed? Use the elation
qv 1 mv ; Hee, q 1 6 coulomb; m 1 3 kg; v 1 m/s Theefoe, V mv q 3 1 1 1 1 1 1 6 5 1 4 volt 5 kv 6 Example : Solution: Thee equal chages Q ae at the vetices of an equilateal tiangle of side A. How much wok is done in binging them close to an equilateal tiangle of side A? In this poblem fist we have to calculate the initial and final potential enegy and then the wok done is final potential enegy minus initial potential enegy.
1 3Q Initial potential enegy 4 A Q and we have thee such combinations) 4 A (fo each combination, the enegy is Final potential enegy Example : Solution: 3Q 4 A Now, wok done W 3Q 4 Theefoe, W 3Q 4 A 3Q A 4 A Two paticles have equal masses of 5 g each and opposite chages of 4 1 5 C and 4 1 5 C. They ae eleased fom est with a sepaation of 1 m between them. Find the speed of paticles when the sepaation is educed to 5 cm. Again in this question we have to do the consevation of mechanical enegy. Initially, thee is only potential enegy but finally thee is both kinetic and potential enegy. Initial state enegy:
Bodies ae at est so kinetic enegy QQ 1 Potential enegy 4 R 9 1 4 1 4 1 14.4 joule Final state enegy: 9 5 5 1 Let us suppose both the chages appoach each othe with individual speed v. (both will have same speed because masses ae same and momentum will be conseved. Since initial momentum is zeo, they will have final momentum also equal to zeo. Since momentum is vecto, they will have equal and opposite momentum) Final kinetic enegy of both 1 mv mv Final potential enegy QQ 4 R 1 9 1 4 1 4 1 8.8 joule Now, 9 5 5.5
initial total enegy final total enegy 14.4 mv 8.8 theefoe, mv 14.4; 14.4 o, v 3 5 1 54 m/s Wok done in otating an Electic Dipole in an Electic Field: If a dipole, placed in a unifom electic field, is otated fom its equilibium position, wok has to be done against the electic field toque. If a small wok dw is done (against the field toque) in otating the dipole though a small angle dө, then dw Toque angula displacement o dw Ө dө pe sin Ө dө Thus, the total wok done against the field toque fo a deflection ' ' fom equilibium is W pesin d pe cos W pe(1 cos Ө) This is the fomula fo the wok done in otating an electic dipole against an electic field though an angleө fom the diection of the field (equilibium position). The wok done in otating fom the position 1 to Ө is given by W pe(cos Ө 1 cos Ө ).
Case (i) If the dipole be otated though 9 fom the diection of the field, then the wok done: W pe(1 cos 9 ) pe (1 ) pe. Case (ii) NOTE: The wok done in otating the dipole though 18 fom the diection of the field W pe (1 cos 18 ) pe [1 ( 1)] pe. The wok done by the field toque on the dipole is W pe (cos Ө cos Ө 1 ) pe(1 cos Ө), when Ө 1, Ө Ө. Potential Enegy of an Electic Dipole in an Electic Field. The potential enegy of an electic dipole in an electic field is defined as the wok done in binging the dipole fom infinity to inside the field. An electic dipole (+q, q) is bought fom infinity to a unifom electic field E in such a way that the dipole moment p is always in the diection of the field. Due to the field E, a foce F qe acts on the chage +q
in the diection of the field, and foce F qe on the chage q in the opposite diection. Hence, in binging the dipole in the field, wok will be done on the chage +q by an extenal agent, while wok will be done by the field itself on the chage q. But, as the dipole is bought fom infinity into the field, the chage q coves l distance moe than the chage +q. Theefoe, the wok done on q will be geate. Hence the 'net' wok done in binging the dipole fom infinity into the field foce on chage ( q) additional distance moved qe l pe. This wok is the potential enegy U of the electic dipole placed in the electic field paallel to it: U pe In this position the electic dipole is in stable equilibium inside the field. On otating the dipole though an angle Ө wok will have to be done on the dipole. This wok is given by W pe(1 cos Ө). This will esult in an incease in the potential enegy of the dipole. Hence, the potential enegy of the dipole in the position Ө will be given by U Ө U + W pe + pe (1 cos Ө). o U Ө pe cos Ө o, U p.e This is the geneal equation of the potential enegy of the electic dipole.
DIPOLE IN A UNIFORM EXTERNAL FIELD Conside a small dipole (+q, q) of length l, placed in unifom electic field of intensity E. Foces of magnitude qe each act on the chage +q, and q as shown in figue. These foces ae equal, paallel and opposite, and hence net foce on the dipole is zeo but these foces constitute a couple. The moment of this couple i.e., toque is Ʈ foce pependicula distance F l sin Ө qe l sin Ө ql E sin Ө o pe sin Ө, whee ql p (electic dipole moment). Ʈ pe sin Ө o, p E If the dipole is placed pependicula to the electic field (Ө 9 o sin Ө 1), then the couple acting on it will be maximum. If this be Ө max, then Ө max PE. If E 1 NC 1, then p Ө max. Hence, electic dipole moment is the toque acting on the dipole placed pependicula to the diection of a unifom electic field of intensity 1 NC 1. Content Builde
Peiod of Dipole in Unifom Electic Field 1. If dipole is paallel o antipaallel to the field (i.e., o 18 ), toque is zeo. Hence dipole is in equilibium position. Conside an electic dipole of dipole moment p making an angle 'Ө' with the field diection. It expeiences a toque pe sin tending to bing it in the diection of field. Theefoe, on being eleased, the dipole oscillates about an axis though its cente of mass and pependicula to the field. If l is the moment of inetia of the dipole about the axis of otation, then the equation of motion is: d l pe sin Ө dt Fo small amplitudes sin Ө - Ө. d Thus, dt Ѡ Ө, pe l whee pe l This is an SHM whose peiod of oscillation is T l. pe. Potential enegy of the electic dipole at an angle Ө with electic field is U P.E PE cos Ө.
3. Wok done is otating the dipole fom angle Ө 1 to Ө is W PE (cos Ө 1 cosө ). Example : Solution: An electic dipole consists of two opposite chages of magnitude. µc each, sepaated by a distance of cm. The dipole is placed in an extenal field of 1 s N/C. What maximum toque does the field exet on the dipole? Toque pe sin Ө Fo maximum value of toque, sin 1 maximum toque pe (. 1 6 1 ) ( 1 5 ) 8 1 4 Nm Motion of a Chaged Paticle In a Unifom Electic Field Along x-axis, u x v, a x, S x x, v x v S ut 1 at x v t Along y-axis, u y,
a y qe, m y y 1 qe t m 1 qe x m v y Ө y Ө x 1 qex mv i.e., path is paabolic. Deflection on the sceen Y (y + D tan Ө) tan Ө v qet qex v mv mv y x So tan Ө qex mv x y tan x which means that the tangent at point P intesect the line AB at its mid point i.e. at distance x fom B. x Y D tan (i) Electic Field at a point on the Axis of a Dipole (end-on position).
Suppose an electic dipole (q, q) of length l is situated in a medium of dielectic constant K. Let P be a point on the axis at a distance mete fom the mid-point O of the dipole. We have to detemine the intensity of the electic field at P. (In figue) Let E and E be the intensities of the electic fields at P due to the chages +q and q of the dipole, espectively. The distance of the point P fom the chage +q is ( l) and the distance fom chage q is ( + l). Theefoe, E 1 1 q 4 K l (away fom chage +q) and E 1 q 4 K l (towads the chage q) The fields E 1 and E, ae along the same line but in opposite diections. Theefoe, the esultant field E at the point P will be E E 1 E 1 q 1 q 4 K 4 K l l 1 1 q 4K l l diected fom q to +q 4q 4 K l
(ii) p o E 4 K l diected fom q to +q. If l is vey small compaed to (l << ), the tem l may be neglected in compaison to. Then, the electic field at the point P due to the dipole is given by 1 p E 4 4 K 1 p 3 4K diected fom q to +q. Fo vacuum (o ai) K 1. 1 p E 3 4 1 p o, E 3 4 The diection of the electic field E is along the axis of the dipole fom the negative chage towads the positive chage, i.e., is in the diection of dipole moment. Electic field at a point on the Equatoial Line of a Dipole (Boadside-on position). In this case, the point P is situated on the ight-bisecto of the dipole AB at a distance mete fom its mid-point. E 1 and E ae the intensities of the electic fields at P due to the chages +q and q of the dipole, espectively. The distance of P fom each chage is
l AP BP l. Theefoe, E 1 1 q 4 K l (in the diection AP ) and E 1 q 4 K l (in the diection PB ). The magnitudes of E 1 and E ae equal (but diections ae diffeent). On esolving E 1 and E into two components, that is, paallel and pependicula to AB, the components pependicula to AB (E 1 sin Ө and E sin Ө) cancel each othe (because they ae equal and opposite) while the components paallel to AB (E 1 cos Ө and E cos Ө), being in the same diection, add up (In Figue). Hence, the esultant electic field at the point P is E E 1 cos Ө + E cos Ө 1 q 1 q cos 4 K 4 K l l q 4 K l cos q 4 K l l o E ql 3 / 4 K l cos
p 4 K l 3 / Fo vacuum (o ai) K 1. E p 3 / 4 K l Fo a small dipole >> l. Theefoe, 1 p E 3 4 1 p o E 3 4 The diection of the electic field is paallel to the axis of the dipole. Content Builde Electic Field due to a Dipole at any Point fo a Small Dipole Suppose the point P on the line making an angle with the axis of a small dipole (l << ). The components of the dipole moment p along and pependicula to OP ae p cos and p sin espectively. Fo the component p cos, the point P is on its axis, so the field at P due to this component is 1 p sin E. 3 4 Fo the component p sin, the point P lies on its equatoial line, so the field at P due to this component is
1 p sin E. 3 4 The esultant field at P is E E E o, E 1 p 4. 3 3cos 1 If E makes an angle Ө with OP, then tan Ө E 1 tan E o, tan tan 1 1 ELECTRIC FLUX (i) (ii) (iii) It is denoted by 'Ө'. It is a scala quantity. It is defined as the total numbe of lines of foce passing nomally though a cuved suface placed in the filed. (iv) It is given by the dot poduct of E and nomal infinitesimal aea ds integated ove a closed suface d E.ds E.ds Eds cos whee Ө angle between electic field and nomal to the aea
(v) (a) if Ө, Ө Eds (maximum) (b) if Ө 9, Ө zeo (vi) Unit: (a) (b) Newton mete / coulomb. Volt mete (vii) Dimension: [M L 3 T 3 A 1 ] (viii) (ix) (x) (xi) (xii) (xiii) (xiv) Flux due to a positive change goes out of the suface while that due to negative change comes into the suface. Flux enteing is taken as negative while flux leaving is taken as positive. Value of electic flux is independent of shape and size of the suface. Flux is associated with all vectos. If only a dipole is pesent in the suface then net flux is zeo. Net flux of a suface kept in a unifom electic field is zeo. Net flux fom a suface is zeo does not imply that intensity of electic field is also zeo. GAUSS'S LAW This law states that electic flux E though any closed suface is equal to 1 times the net chage 'q' enclosed by the suface i.e., E E.ds
NOTE: q The closed suface can be hypothetical and then it is called a Gaussian suface. If the closed suface enclosed a numbe of chages q 1, q... q n etc. then E.ds q q 1 q...qn Flux is Note (i) (ii) (iii) (iv) Independent of distances between chages inside the suface and thei distibution. Independent of shape, size and natue of suface. Dependent on chages enclosed by suface, thei natue and on the medium. Net flux due to a chage outside the suface will be zeo. (v) If Q, then but it is not necessay that E. (vi) (vii) Gauss law is valid only fo the vecto fields which obey invese squae law. Gauss's and coulomb's law ae compaable.
(i) A chage q is placed at the cente of a cube, then (ii) (a) Total flux though cube q (b) Flux though each suface q 6 A chage q is placed at the cente of a face of a cube, then total q flux though cube How A second cube can be assumed adjacent to the fist cube total flux though q q both cubes, So flux though each cube (iii) Now, q is placed at a cone then the flux will be q. 8 Gauss's Law Example: Solution: A hemispheical suface of adius R is kept in a unifom electic field E such that E is paallel to the axis of hemi-sphee, Net flux fom the suface will be E.ds E. R (E) (Aea of suface pependicula to E) E. R.
Example: NOTE: A ectangula suface of length 4m and beadth m is kept in an electic field of N/c. Angle between the suface and electic field is 3. What is flux thought this suface? Angle between suface and E is given to be 3. This is not the 'Ө' used in ou fomula 'Ө' is the angle between nomal to suface and E. So hee Ө 9 3 6. Solution: Ө EA cos Ө 8 cos 6 8 V-m Example: Solution: In the following, find out the emeging electic flux though S 1 and S whee [q 1 1µc, q µc, q 3 3µc] q 1 1 6 1 8.85 1 1 1.13 1 5 V.m
Example: Solution: Example: A chage 'q' is placed at the cente of a cube of side 'a'. If the total flux passing though cube and its each suface be ф 1 and ф espectively then ф 1 : ф will be (A) 1 : 6 (B) 6 : 1 (C) 1 : 6a (D) 6a : 1 (B) When q is placed at the cente of cube then total flux passing though cube is q 1 and flux though each suface is q 6 6 : 1 : 1 If chages q and q ae placed at the cente of face and at the cone, of a cube. Then total flux though cube will be (A) q (B) q
(C) q 6 (D) q 8 Solution: (A) Flux though cube, when q is placed at the cente face, is q 8 q, 4 Total flux 1 Example: Solution: q q 4 4 1q Flux enteing a closed suface is V-m. Flux leaving that suface is 8 V-m. Find the chage inside suface. Net flux ф out ф in ф (8 ) 6 V-m q q (6) (8.85 1 1 ).53 µc
APPLICATION OF GAUSS'S LAW Electic field due to a chaged conducting sphee/ Hollow conducting o insulating sphee. (i) In all the thee type of sphees, chage esides only on the oute suface of the sphee in ode to emain in minimum potential enegy state. Case 1: OP > R E q q 4 1 R ( suface chage density) Case : R E Case 3: < R E
i.e. At point inteio to a conducting o a hollow sphee, electic field intensity is zeo. (iii) (iv) Fo points outside the sphee, it behaves like all the chage is pesent at the cente. Intensity of electic field is maximum at the suface. Imp. (v) Electic field at the suface is always pependicula to the suface. (vi) Fo points, nea the suface of the conducto, E pependicula to the suface. (vii) Gaphically, Electic potential Case 1: < R V in 1 Q 4 R Case : R V suface 1 Q 4 R Case 3: > R
1 Q V out 4 (i) Fo points inteio to a conducting o a hollow sphee, potential is same eveywhee and equal to the potential at the suface. (iii) at, V NOTE: Hee, we see that E inside the sphee is zeo but V. So E does not imply V. This pesents a good example fo it. Similaly V does not imply E. Electic field due to solid insulating sphee A chage given to a solid insulating sphee is distibuted equally thoughout its volume Electic Field Case 1: > R (point is outside the sphee) Case : E 1 Q 4 R (point is at the suface) E 1 Q 4 R Case 3: E max E suface
< R (point is inside the sphee) (i) (ii) (iii) (iv) E Ein 3 1 Q 3 4 R at, E Gaphically Again, fo points outside the sphee, it behaves as all the chage is pesent at the cente. Fo points outside, it obeys invese squae law. Intensity of electic field at infinity is zeo. 1 Q (v) Intensity at the suface is maximum and is equal to. 4 R (vi) (vii) Again, it is pependicula to the suface at the suface. Intensity is zeo at the cente and fo points inside the sphee, it is diectly popotional to distance of the point fom the cente. Electic Potential Case 1: > R V out Case : R 1 Q 4
Case 3: < R 1 Q V suface 4 R V in 1 Q 3R 4 R 3 1 Q V cente (Imp) 4 R 3 (i) (ii) (iii) (iv) V cente Gaphically 3 V suface Again, E cente, but V cente Electic potential at infinity is zeo. Electic potential is maximum at the cente. Example: A solid insulating sphee of adius R is given a chage. If inside the sphee at a point the potential is 1.5 times that of the potential at the suface, this point will be (A) At the cente (B) (C) At distance 3 R fom the cente Potential will be same inside and on the suface of sphee, so given infomation is inadequate.
(D) Insulating bodies cannot be given chage. Solution: (A) Potential at the cente of insulating sphee is given by V in 1 Q 3R 4 R and on the suface, V suface given that, 1 Q 4 R 3 () (1) V in 3 V suface Q 3R 3Q 3 R R Hence the point will be at the cente. Example: Solution: Two concentic sphees of adii & R ( < R) ae given the chages q and Q espectively. Find the potential diffeence between two sphees. Potential at the inne sphee potential due to inne + potential due to oute sphee 1 q 1 Q V1 4 4 R
(potential at points inside is same eveywhee and is equal to potential at the suface). Potential at oute sphee V potential due to inne + potential due to oute sphee 1 q 1 Q 4 R 4 R potential diffeence V 1 V 1 q q 4 R NOTE: Example: q 1 1 V 4 R Hee, we see that 'V' depends only on the chage of inne sphee. In the following fig, of chaged sphees A, B & C whose chage densities ae σ, σ & σ and adii a, b & c espectively what will be the value of V A & V B. Solution: 4a 4b 4c V A k k k a b c a b c
q A 4a 4b 4c V B k b b c a b c b Ex. Figue shows a unifomly chaged sphee of adius R and total chage Q. A point chage q is situated outside the sphee at a distance fom cente of sphee. Find out the following: (i) Foce acting on the point chage q due to the sphee. (ii) Foce acting on the sphee due to the point chage. Sol. (i) Electic field at the position of point chage E KQ ˆ so, KqQ ˆ
F KqQ (ii) Since we know that evey action has equal opposite eaction so F KqQ ˆ sphee F KqQ sphee Ex. Figue shows a unifomly chaged thin sphee of total chage Q and adius R. A point chage q is also situated at the cente of the sphee. Find out the following: (i) Foce on chage q (ii) Electic field intensity at A. (iii) Electic field intensity at B. Sol. (i) Electic field at the cente of the unifomly chaged hollow sphee
So foce on chage q (ii) Electic field at A E E E A Sphee q Kq ; CA E due to sphee, because point lies inside the chaged hollow sphee. (iii) Electic field E B at point B E Sphee E q KQ Kqˆ K Q q ˆ ; CB
Hee we can also assume that the total chage of sphee is concentated at the cente, fo calculation of electic field at B. Ex. Two concentic unifomly chaged spheical shells of adius R 1 and R (R > R 1 ) have total chages Q 1 and Q espectively Deive an expession of electic field as a function of fo following positions. (i) < R 1 (ii) R 1 < < R (iii) > R Sol. (i) fo < R 1, theefoe point lies inside both the sphees Enet Einne Eoute (ii) fo R 1 < < R, theefoe point lies outside inne sphee but inside oute sphee : E net E inne + E oute
KQ1 ˆ KQ1 ˆ (iii) fo > R point lies outside inne as well as oute sphee theefoe. E net E inne + E oute KQ KQ ˆ 1 ˆ K Q1 Q ˆ Ex. A solid non conducting sphee of adius R and unifom volume chage density has its cente at oigin. Find out electic field intensity in vecto fom at following positions : (i) R,,
(ii) R R,, (iii) (R, R, ) Sol. (i) at R,, : Distance of point fom cente R R R, so point lies inside the sphee so E 3 R î 3
(ii) At R R,, ; distance of point fom cente R R R R so point lies at the suface of sphee, theefoe E KQ 3 R 4 3 K R 3 R ˆ R i ˆj 3 R R ˆ R i ˆj 3 (iii) The point is outside the sphee KQ so, E 3
4 3 K R 3 Ri ˆ Rj ˆ 3 R Ri ˆ Rj ˆ 6 (b) Electic Field inside a Cavity of Non-conducting Chaged Body: Conside the sphee shown in figue chaged unifomly with chage density coul/m 3. Inside the sphee a spheical cavity is ceated with cente at C. Now we wish to find electic field stength inside the cavity. Fo this we conside a point P in the cavity at a position vecto x fom the cente of sphee and at a position vecto y fom the cente of cavity as shown. If E 1, be the electic field stength at P due to the complete chage of the sphee (inside cavity also) then we known electic field stength inside a unifomly chaged sphee is given as
x E 1 ; 3 E y 3 Now the electic field due to the chaged sphee in the cavity at point P can be given as E E E net 1 [As now chage of cavity is emoved] a 3 [As x y a] This shows that the net electic field inside the cavity is unifom and in the diection of a i.e. along the line joining the cente of sphees and cavity.
Ex. Two conducting hollow spheical shells of adii R and R cay chages Q and 3Q espectively. How much chage will flow into the eath if Inne shell is gounded? Sol. When inne shell is gounded to the Eath then the potential of inne shell will become zeo because potential of the Eath is taken to be zeo. Kx K3Q R R x 3Q, the chage that has inceased 3Q Q Q hence chage flows into the Eath
Q Ex. An isolated conducting sphee of chage Q and adius R is connected to a simila unchaged sphee (kept at a lage distance) by using a high esistance wie. Afte a long time what is the amount of heat loss? Sol. When two conducting sphees of equal adius ae connected chage is equally distibuted on them. So we can say that heat loss of system H U 1 U 1 Q Q Q 4 4 8 R 8 R 8 R Q 16 R Ex. The two conducting spheical shells ae joined by a conducting wie and cut afte some time when chage stops flowing. Find out the chage on each sphee afte that.
Sol. Afte cutting the wie, the potential of both the shells is equal Thus, potential of inne shell V in Kx R K Q x R k x Q R and potential of oute shell V out Kx R K Q x R KQ R As V out V in KQ K x Q R R Q x Q x
So chage on inne spheical shell and oute spheical shell Q. Ex. Find chage no each spheical shell afte joining the inne most shell and oute most shell by a conducting wie. Also find chages on each suface. Sol. Let the chage on the innemost sphee be x. Finally potential of shell 1 Potential of shell 3 Kx K Q K 6Q x R R 3R Kx K Q K 6Q x 3R 3R 3R 3x 3Q + 6Q x 4Q; x Q; x Q
Ex. Thee ae 4 concentic shells A, B, C and D of adius of a, a, 3a, 4a espectively. Shells B and D ae given chages +q and q espectively. Shell C in now eathed. Find the potential diffeence V A V C. Sol. Let shell C acquies chage 'q' which will be such that final potential of C is zeo. kq kq' kq VC 3a 3a 4a kq kq' kq 3a 3a 4a 1 1 q' 3q 4 3 q' q 4 As V C V A V C V A Now calculating V A we get
q k kq 4 kq V A a 3a 4a V A kq 6a o V V A C kq 6a Ex. Figue shows thee lage metallic plates with chages Q, 3Q and Q espectively. Detemine the final chages on all the sufaces. Sol. We assume that chage on suface is x. Following consevation of chage, we see that sufaces 1 has chage ( Q x). The electic field inside the metal plate is zeo so fields at P is zeo. Resultant field at P E P Q x x 3Q Q A A
Q x x 4Q x 5Q We see that chages on the facing sufaces of the plates ae of equal magnitude and opposite sign. This can be in geneal poved by gauss theoem also. Remembe this it is impotant esult. Thus the final chage distibution on all the suface is as shown in figue: (A) TOTAL ELECTROSTATIC ENERGY OF A SYSTEM OF CHARGES: Total electostatic potential enegy of system of chages can be given as U self enegy of all chaged bodies + Inteaction enegy of all pais of chaged bodies. Let us conside some cases to undestand this concept. Figue shows two unifomly chaged non conducting sphees of adius R 1 and R and