CHAPTER 25 ELECTRIC POTENTIAL

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1 CHPTE 5 ELECTIC POTENTIL Potential Diffeence and Electic Potential Conside a chaged paticle of chage in a egion of an electic field E. This filed exets an electic foce on the paticle given by F=E. When the paticle moves fom initial point i to final point f this foce does wok on it accoding to W = f i E dl Since the electic foce is consevative, we can associate a potential enegy with this foce accoding to o U = W U f U i = f i E dl Let us now define the electic potential as the electic potential enegy pe unit chage, i.e., U = The electic potential diffeence between the points i and f is then f i f = E dl i

2 The potential diffeence f i is defined as the wok euied to move a positive unit chage fom point i to point f. Since it is only the change in potential between two points that has physical sense, it is often convenient to choose a efeence point fo zeo potential. This efeence point is usually chosen to be at infinity, i.e., = 0. Letting the point i to be and designate point f as point p, the electic potential at any point p becomes p p = E dl That is, the electic potential at any point p which can be defined as the wok euied to bing a positive unit chage fom infinity to that point. In this sense the potential at a point is the potential diffeence between that point and a point at infinity. The electic potential is a scala uantity. The SI unit of is joules pe coulomb (J/C), usually epesented by a special unit called volt (), i.e., = J/C It follows that the electic field has a unit of volt pe mete (/m) with N/C = /m Example 5. poton is eleased fom est in a unifom e.f. that has a magnitude of /m. The poton undegoes a displacement of 0.5 m in the diection of E. a) Find b) Find U U c) Find the speed of the poton at point.

3 Solution: = = E dl = Edl This follows fom the fact that E and dl ae paallel. Since E is unifom it can be taken out fom the integal sign, giving = E dl = Ed 4 4 = ( )( 0.5) = Fom the minus sign we conclude that >. E 9 4 U = = (.6 0 )( ) b) = J c) K U = 0 ( ) ( ) = 0 mv v = 5 ( ) =.8 0 m/s

4 ELECTIC POTENTIL DUE TO POINT CHGES dl Conside an isolated point chage. To find the electic potential due to this chage at a point, we have = E dl The electic field due to the chage at a distance is E =k ˆ, whee ˆ is a unit vecto diected adially outwad fom. Theefoe = k ˆ dl ut ˆ d l = d = k = k k

5 The fist tem of the ight hand side epesents the potential at point ( ) and the second tem epesents the potential at point ( ). The electic potential at a point a distance fom a point chage is then obtained if, that is, = k s it is clea fom the above euation, is positive fo positive and negative fo negative. The electic potential due to a goup of point chages at a point is the algebaic sum of the electic potentials due to each chage individually. That is, fo a goup of N point chages we have = N i k i i whee i is the distance fom the ith chage i to the point in uestion. Do not foget that the sum is algebaic sum and not vecto sum like that used to calculate the electic filed due to a goup of point chages. This fact gives an impotant advantage of potential ove electic field. Example 5. Two point chages of =.0 µc, = 6.0 µc, ae aanged as shown. Find a) the electic potential at the point p. b) the change in potential enegy of the system plus a chage =.0 µc that moves fom to point P. = 6.0 µc.0 m =.0 µc 0.5 m 4.0 m θ p Solution:

6 a) = k 9 6 = = b) The change in P.E is eual to the wok euied, by an extenal agent, to bing fom to point P, i.e., U = W = = P 6 (.0 0 )( ) =.89 0 J POTENTIL ENEGY OF SYSTEM OF POINT CHGES Let us calculate the wok euied to assemble a system of thee point chages. To do so we have to calculate the wok euied to bing, fom infinity, each chage one by one. s the potential at a point is eual to the wok euied to bing a positive unit chage fom infinity to that point. Theefoe, the wok euied to bing a chage fom infinity to a point must eual to the potential at that point multiplied by, i.e., W = p p

7 Now, no wok is euied to place the fist chage at a given position because thee is no electic potential at that position. Next we place a second chage at a position fom. This euies a wok whee is the potential at the location of due to. If we denote this wok by W with W =0 denotes the wok euied to place, then we have W = k Next we place at a position fom and fom. We now must do wok given as, whee is the potential at the location of due both and, i.e., W = k The total wok euied to assemble a system of thee chages is then W = W W W This wok is stoed as an electostatic enegy in the system, so we wite U = = W k The potential enegy of a system of N chages can be calculated in a simila fashion. Fo 4chages, fo example we have = = U W k

8 Fo a simple system of only two point chages the potential enegy is given by U = k Note that if the chages ae of the same sign U is positive which means that the foce is epulsive, as expected. If the chages ae of the opposite sign U is negative which means that the foce is attactive, which is the case. Example Find the potential enegy of the system given in the pevious example. Solution: U = k (.0)( 6.0) (.0)(.0) 0 = (.0)( 6.0) 0 5 = J

9 E. POTENTIL DUE TO CONTINUOUS CHGE DISTIUTION If a chage is continuously distibuted in a given egion, two methods will be followed: If the chage distibution is well known we can poceed as we do fo calculating the electic field discussed: choosing an element of chage d, finding the potential d due to this element, and then integating ove the entie chage distibution. This method is well illustated in finding the potential du to a line of chage, chaged ing, and chaged disk. If the electic field of the chage distibution can be calculated easily (Using Gauss law fo instant), we can use the Euation p p = E dl This method is useful fo cases whee Gauss law can be applied, i.e., fo chage distibution with high degee of symmety. Example 5.7 wie of length L has a chage Q unifomly distibuted. Find the electic potential at a point along its axis and a distance d fom one end. a y Solution: We divide the wie into small elements each of length dx and chage d. Since each element can be consideed as a point chage, the electic potential d due to one of these elements a distance x fom p is x L d dx x d d = k = k x d a

10 Since we ae dealing with a scala uantity we can integate the above expession diectly to get the potential due to the whole wie. The esult is = k x d a = kλ L 0 x dx a whee we have substituted fo d as λdx. Pefoming the integal we get L = kλ ln L a a kq L L a = ln L a Example 5.5 ing of adius has a unifom chage distibution of magnitude Q. Calculate the electic potential along the axis of the ing at a point P lying a distance D fom the cente of the ing. Solution: The ing is divided into small elements each of length ds and chage d. The electic potential due to one element which can be consideed as a point chage is d D θ P d d= k = k d ( D )

11 Integating we get Q k = d = D 0 Qk D t the cente of the ing D=0 and the electic potential is then Q = k Example 5.6 disk of adius has a unifom chage density σ. Calculate the electic potential along the axis of the disk at a point p lying a distance D fom the cente of the disk. p Solution: We divide the disk into thin ings each of thickness d. Fom the esult of the pevious example, the potential due to one element with adius is d D d= k d ( D ) Integating the last esult and noting that d= σd= σ we get E= kπσ 0 ( π d ) d ( D ) = π kσ D D = kπσ D 0

12 Example 5.8 n insulating solid sphee of adius has a chage Q unifomly distibuted within its volume. Find the electic potential at a point a) outside the sphee >, b) inside the sphee <. Q Solution: We define E and E to be the electic fields in the >, and < egions, espectively. Using the esult of the pevious Example we have E Q = ˆ > 4πε o E Q = ˆ < 4πε o a) Fo outside the sphee >, we have = E d Substituting fo E by E and knowing that d = dˆ we get d = kq Q = k = kq Note that this esult is identical to that of a point chage. This means that the poblem can be teated as the entie chage was concentated at the cente of the sphee. b) Fo inside the sphee <, we have

13 = E d ut in this case the ange fom to the point in uestion enclose two egions with diffeent electic fields: The outside egion ( > ) which extend fom to, and the inside egion ( < ) which extend fom to. Theefoe, we have to divide the integation into two pats as = E d E d Substituting fo E and E we obtain Q d Q = 4πε 4πε o o d Pefoming the integals and eaange we get Q 8πε o = Note that the two expessions of and give the same esult fo = (at the suface) which is Q = 4πε o

14 ELECTIC POTENTIL ND CONDUCTOS In the pevious chapte we have mentioned some popeties of a conducto in electostatic euilibium with the help of Gauss law. Some of these popeties ae: The electic field inside the conducto is zeo. The electic field on the suface of the conducto is pependicula to the suface. Now we have f i f = E dl i If the initial and the final points lie inside a conducto and using the fist popety we conclude that the electic potential is constant inside the conducto. Futhemoe, since E is always nomal to dl on the suface, the electic potential is also constant on the suface of the conducto. nothe impotant popety of a conducto thus follow The electic potential inside any conducto in euilibium is constant and eual to the potential on its suface. To pove this popety explicitly, let us conside a conducting sphee of adius and chage Q. The electic potential at a point inside the sphee is = ut E in =0 and E d = E out E out d E in d Q = ˆ, so we get 4πε o kq

15 = Q 4πε o which is the value of the potential at the suface of the conducto. The figue shows the vaiation of as a function of the adial distance fo a conducting sphee of adius. If two o moe conducting objects ae connected by a conducting wie, the conductos ae no longe sepaate but can be consideed as a single conducto. This means that the electic chages will tansfe fom the conducto of highe potential to that of lowe potential until the euilibium condition is achieved. Theefoe, if two o moe conductos ae connected and euilibium is achieved, they must be at the same electic potential. In analogue with the electic filed lines, the electic potential can be epesented by euipotential sufaces. suface with all its points ae at the same electic potential is called euipotential suface. Fom this definition, it follows that no wok is done by the electic field in moving a chaged paticle between two points on the same euipotential suface. This means that the electic filed lines must be pependicula to the euipotential sufaces. The suface of any conducto foms an euipotential suface. Example 5.9 Two conducting chaged sphees with adii and ae sepaated by a distance much lage than the adius of eithe sphee. The two sphees ae connected by a conducting wie. The chages on the sphees in euilibium ae and. a) Find the atio of the magnitudes of the e. fields at the sufaces of the sphees, and the atio of the chage densities on the sufaces of the sphees. Solution: Since the sphees ae connected by a conducting wie, the potential is the same fo both sphees, i.e., =

16 K = K Fom which it follows that = The chages ae distibuted ove the suface of the sphees, so we have σ = and 4π σ = 4π Theefoe σ σ = Substituting fo the atio given above we get σ σ = It is clea fom this esult that the chage density is geatest on the small sphee as expected. σ Since the e. field on the suface of a conducto is eual to we get ε o E E = σ σ = That is, the field is moe intense nea the smalle sphee.

CHAPTER 10 ELECTRIC POTENTIAL AND CAPACITANCE

CHAPTER 10 ELECTRIC POTENTIAL AND CAPACITANCE CHAPTER 0 ELECTRIC POTENTIAL AND CAPACITANCE ELECTRIC POTENTIAL AND CAPACITANCE 7 0. ELECTRIC POTENTIAL ENERGY Conside a chaged paticle of chage in a egion of an electic field E. This filed exets an electic