Target Boards, JEE Main & Advanced (IIT), NEET Physics Gauss Law. H. O. D. Physics, Concept Bokaro Centre P. K. Bharti

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1 Page 1 CONCPT: JB-, Nea Jitenda Cinema, City Cente, Bokao Gauss Law Autho: Panjal K. Bhati (IIT Khaagpu) Mb: Taget Boads, J Main & Advanced (IIT), NT 15 Physics Gauss Law Autho: Panjal K. Bhati (B. Tech., IIT Khaagpu) H. O. D. Physics, Concept Bokao Cente 7 P. K. Bhati All ights eseved

2 Page CONCPT: JB-, Nea Jitenda Cinema, City Cente, Bokao Gauss Law Autho: Panjal K. Bhati (IIT Khaagpu) Mb: Peeuisite Vectos Newton s Laws Coulomb s Law lectic Field Please evise these topics befoe studying Gauss s Law Intoduction We ae going to intoduce the concept of Gauss s Law which can be used fo symmetical situation with utmost ease. Using Gauss Law we can we find electic field in just two-thee steps fo a symmetical situation. Concept of aea vecto and flux is an impotant pat of Gauss s Law which we will intoduce befoe intoducing Gauss s Law. Aea Vecto Let us stat with an example of suae suface of aea A. Aea vecto A coesponding to this suae is a vecto pependicula to its suface having magnitude eual to its aea A. Thus, A = A We can find out total suface aea (in scala fom) of an iegula suface using integation. Total suface aea is given by A= = d Aea Vecto (Summay) Aea vecto of a flat suface is a vecto pependicula to suface having magnitude eual to its aea A. Aea vecto is taken in outwad nomal diection fo a closed suface. We can find out total suface aea (in scala fom) of an iegula suface using integation. uick xecise 1 1. Conside the given cube of edge m in given coodinate system. Wite aea vectos of all six faces in unit vecto notation. z D G C y O F A B x Again let us see an example of a cube of edge a. Clealy, suface aea of each face is a. Thus, a cube have six aea vectos coesponding to each side as shown in the figue. Clealy, ATop = ABottom = ALeft = Aight = AFont = ABack = a. One impotant thing to note hee is that each aea vecto is taken in the outwad nomal diection. These wee standad figues, whee we can define aea vecto in (outwad fo a closed suface) nomal diection. All faces wee flat. What if we get an iegula suface? Fo the case of an iegula suface we divide the suface into infinitesimally small aeas. Let us conside such a small aea to be. Then, aea vecto coesponding to this small suface is in the outwad nomal diection as shown in the figue.. Suppose a unifom electic field exists in a egion along ve x-axis. Coss section of a flat suface having aea m at an angle 45 o with x-axis is shown in figue. Find aea vecto in unit vecto notation. y O 45 o Answes: 1. Face ABF: 9 mi, face OCDG: 9 mi, Face BCD: 9 m j, face OAFG: 9 m j, Face DFG: 9 mk, face OABC: 9mk. i j m ( ) x

3 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: lectic Flux xample 1 Physical meaning of Flux: oughly speaking, we can pictue flux φ in tems of numbe of electic field lines passing nomally though an aea A. Inceasing the aea means that moe lines of pass though the aea, inceasing the flux. Stonge field means moe closely spaced lines of and theefoe moe lines pe unit aea, so again the flux inceases. Definition (Fo Boads): The total numbe of electic field lines passing nomally though an aea placed inside an electic field is called electic flux. xpession fo flux: lectic Flux though a flat suface of aea A, when unifom electic field passes though it in any diection is given by φ = A. (lectic flux though a flat aea in a unifom field) Find the net flux though a closed cylinde of length l and adius in a unifom electic field paallel to its axis. Solution: The closed cylinde is composed of thee sufaces: left cicula suface, ight cicula suface and a cuved suface. Thus, net flux though the cylinde, φ net = (flux though ight cicula suface, φ left ) (flux though left cicula suface, φ ight ) (flux though cuved suface, φ Cuved ). i.e., φ net = φ Left φ ight φ Cuved Let us calculate flux though each suface. Flux though ight cicula suface: (1) Conditions fo this expession: a) Suface is flat & b) lectic field is unifom Flux fo geneal case: We have aleady defined flux though a flat suface in a unifom electic field. What happens if the electic field isn't unifom but vaies fom point to point ove the aea A? O what if A is pat of a cuved suface? We divide suface aea A into many small elements, with aea vecto. We calculate the electic flux. though each element and integate the esults to obtain the total flux: φ = (lectic flux fo geneal case) Fo a completely closed suface we can use closed. Cicle ove integal sign is integal in place of used to denote integation ove a closed suface. φ = (lectic flux though a closed suface) Aea vecto Aight fo ight suface must be taken in outwad nomal diection, i.e., towads ight. As the ight suface is flat and field is unifom, we can use φ = A. Thus, flux though ight cicula suface φ =. Aight = A cos = π...() ight ight Flux though left cicula suface: Aea vecto Aleft fo left suface must be taken in outwad nomal diection, i.e., towads left. Thus, flux though left cicula suface φleft =. Aleft = A cos18 = π...() left Flux though cuved suface: As this suface is not flat we can t use φ = A. We will have to use, φ =.. Let us take a small element of aea on the cuved suface. Clealy, angle between and is 9. Thus, flux though this small aea, o dφ =. d A = cos 9 =. If we take any such element on the cuved suface, aea vecto will be pependicula to. He, flux due to all such elements on the cuved suface =. Page CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

4 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: He, net flux though cuved suface, φ Cuved =. =...(4) Using ens. (1), (), () & (4), we get, φ net = φ left φ ight φ Cuved φ net = He, flux though a closed cylinde in a unifom electic field = zeo. DISCUSSION: We have seen fom above example that flux though a closed cylinde in a unifom electic field is zeo. This is tue fo all closed suface completely inside a unifom electic field whethe it is cylinde o not. We know that the net flux though the suface is popotional to the net numbe of lines leaving the suface, whee the net numbe means the numbe leaving the suface minus the numbe enteing the suface. Unifom electic field means that electic field lines ae paallel and spacing between any two field line is same. He, if a closed suface is placed in a egion of unifom electic field, the net numbe of lines leaving the suface is zeo, since same numbe of field lines leave as that which ente. Thus, flux though a closed suface in a unifom electic field is zeo. 4. An electic field in a egion is given by 1 = i NC fo x > 1 and = i NC fo x < A ight cicula cylinde of length cm and adius 5 cm has its cente at the oigin and its axis along the x-axis so that one face is at x = 1 cm and othe is at x = 1 cm. i. What is the flux though each flat face? ii. What is the flux though the side of the cylinde? iii. What is the net outwad flux though the cylinde? 5. In a egion of the space, the electic field is in the x- diection and popotional to x, ie.., = x. Find the flux though the shaded aea. (IIT-J 11) Solutions: uick xecise NCT uestions Conside a unifom electic field = 1 i N/C (a) What is the flux of this field though a suae of 1 cm on a side whose plane is paallel to the yz plane? (b) What is the flux though the same suae if the nomal to its plane makes a 6 angle with the x-axis? What is the net flux of the unifom electic field of xecise 1.15 though a cube of side cm oiented so that its faces ae paallel to the coodinate planes? Moe uestions fo pactice 1. A cicula plane of adius 1 cm is placed in a unifom electic field of NC -1, making an angle of 6 with the field. Calculate electic flux though the sheet.. If = 6 i j 4 k N/C, calculate the electic flux ( ) though a suface of aea m in Y-Z plane.. In a egion of the space, the electic field is in the x- diection and popotional to x, i. e., = xi. Conside an imaginay cubical volume of edge a, with its edges paallel to the axes of coodinates. Find the flux though this volume. NCT uestions: (a) Nm C -1 (a) 15 Nm C zeo Moe uestions fo pactice Nm C Nm C -1. a 4. i Nm C -1 though each flat face ii. zeo iii..14 Nm C a Page 4 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

5 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: Gauss Law In this section we descibe a geneal elationship between the net electic flux though a closed suface (often called a Gaussian suface) and the chage losed by the suface. Definition: Gauss' law states that the total electic flux 1 though any closed suface is times total chage losed by that suface. Thus, φ = ( Gauss' Law ) Impotant points to note about Gauss s Law: 1. φ = electic flux though closed suface because of electic field ceated by all chages (inside and outside the Gaussian suface). = total chages which ae inside the Gaussian suface. Using definition of flux though a closed suface, we can wite Gauss s Law as. = ( Gauss' Law ) Altenate definition of Gauss s Law: The suface integal of 1 electic field though any closed suface is times total chage losed by that suface. Gauss's law is an altenative to Coulomb's law. Gauss s law is moe fundamental law than Coulomb s Law. It is vey useful in symmetic situations. Deivation of Gauss s Law Let us conside a point positive chage. Let us conside the total flux though a sphee of adius, which loses this point chage. Now, let us conside a small suface of aea and aea vecto on the suface of the sphee. Clealy, angle between and is. Thus, flux though this small aea, dφ =. d A = cos =. If we take any such element on the cuved suface, aea vecto will be paallel to. He, net flux due to all such elements on the Gaussian suface φ =. d A = φ =...(1) We have taken outside integal because magnitude of electic field is same eveywhee on the suface of the Gaussian suface. Now, φ = = ( 4π ) = ( 4 π ) 4π = suface aea ofspheical suface = 4 π & = 4π φ = This poves Gauss s law. Constuction of Gaussian Suface Point chage: We should use symmety to constuct a Gaussian suface. The most appopiate Gaussian suface fo a point chage is a sphee. The point chage lies at cente of the sphee. The most appopiate Gaussian suface fo an infinite line chage is a cylinde. The line chage lies on the axis of the cylinde. An appopiate Gaussian suface fo an infinite plane chage is a cylinde. Axis should be pependicula to the suface. Point chage λ Infinite Line chage APPLICATION OF GAUSS S LAW Step by step method to find electic field using Gauss s law: 1. Gauss s Law is geneally used fo symmetical situation. Fist constuct a closed Gaussian suface using symmetical situation.. Daw electic field lines.. Take a small aea vecto pependicula to suface at one o moe points accoding to situation. 4. Find flux though this closed suface using φ =. Most of the cases we will find that and ae eithe paallel o anti-paallel fo an element of aea. Thus, we will get. d A = when & ae paallel and. d A = when & ae anti-paallel. 6. Find the total chage inside the Gaussian suface. Finally use Gauss s Law φ =. = and get electic field magnitude. σ Infinite Suface chage Page 5 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

6 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: Application of gauss s law: Field due to a point chage Let us conside a point positive chage. Suppose we have to find out electic field at point P located at a distance fom this chage using Gauss s Law. The Gaussian suface is in the fom of a sphee. The adius of spheical Gaussian suface should be, because we ae inteested to find out electic field at a distance fom the chage. We know that electic field due to a point positive chage in the adial outwad diection. Also the magnitude of electic field at any point on the Gaussian suface is same because evey point is at same distance fom the point chage. Now, let us conside a small suface of aea with aea vecto on the suface of the sphee. Clealy, angle between and is at each point on the Gaussian suface. Thus, flux though this small aea, dφ =. d A = cos =. If we take any such element on the cuved suface, aea vecto will be paallel to. He, flux due to all such elements on the Gaussian suface φ =. d A = φ =...(1) Now, = suface aea of Gaussian suface = 4ππ He, en (1) becomes φ = = 4π Now we ae eady to use Gauss s Law, which is: φ = 4π = = 4π Clealy this expession fo electic field due to a point chage is same as that of when deived by Coulomb s Law. lectic field due to an infinite line chage Conside a long line chage with a linea chage density λ. We have to calculate the electic field at a point P which is at a distance fom the line chage. The electic field must be along the adial outwad diection. We daw a cylinde of length l passing though P and coaxial with the line chage as a Gaussian suface. Flux though cuved pat: The electic field at all the points on the cuved suface have the same magnitude as that at P. Also, the diection of the field at any point on the cuved suface is nomal to the line chage and he nomal to the cylindical suface element thee. The flux though the cuved pat is, theefoe,. d S = ds cos = ds = ds = π l. Flux though flat pats: Now, conside the flat pats of the Gaussian suface. The field and the aea-vecto make an angle of 9 o with each othe so that ds. = on these pats. Net flux though Gaussian suface: Cleay, net total flux though the closed Gaussian suface φ =. d S = πl.... ( i) Gauss s Law: The chage losed in the Gaussian suface is = λl as a length l of the line chage is inside the closed suface. Using gauss s law and (i), we have, ( λl) φ = πl = λ = (lectic field due to a line chage) π This is the field at a distance fom the line. It is diected away fom the line if the chage is positive and towads the line if the chage is negative. Vaiation of electic field with distance fom line chage is shown in given gaph. Clealy, electic field deceases with distance fom line chage. P Page 6 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

7 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: lectic Field due to a Unifomly Chaged Spheical Shell Suppose a total chage is unifomly distibuted in a nonconducting spheical shell of adius and we ae euied to find the electic field at a point P which is at a distance fom the cente of the chage distibution. The electic field due to a unifomly chaged spheical shell at an intenal point is zeo. Vaiation of electic field with distance fom the cente is shown in following gaph: Field at an outside point: Fo a point P outside the chage distibution we have >. Daw a spheical Gaussian P suface passing though the point S S P and concentic with the chage distibution. The electic field is adial by symmety and if is positive the field is outside. Also, its magnitude at all the points of the Gaussian suface must be eual. As the field is nomal to the suface element eveywhee,. d S = ds fo each element. The flux of the electic field though this closed suface is φ =. d S = ds = ds = 4 π. Using Gauss s Law, we get 4π = =. 4π The electic field due to a unifomly chaged spheical shell at a point outside it, is identical with the field due to an eual point chage placed at the cente. Field at an intenal point: Suppose, we wish to find the electic field at a point P inside the spheical chage distibution. We daw a spheical suface of adius passing though P and concentic with the given chage distibution. Clealy, the entie chage is on the suface of the spheical shell. Theefoe, thee is no chage inside ou Gaussian P suface, as it is inside the shell. He, =. Using Gauss s law, φ = ds. = = lectic Field due to a Unifomly Chaged Solid Sphee Suppose a total chage is unifomly distibuted in a nonconducting spheical volume of adius and we ae euied to find the electic field at a point P which is at a distance fom the cente of the chage distibution. Field at an outside point: Fo a point P outside the chage distibution we have >. Daw a spheical Gaussian P suface passing though the point S S P and concentic with the chage distibution. The electic field is adial by symmety and if is positive the field is outside. Also, its magnitude at all the points of the Gaussian suface must be eual. As the field is nomal to the suface element eveywhee,. d S = ds fo each element. The flux of the electic field though this closed suface is φ =. d S = ds = ds = 4 π. Using Gauss s Law, we get 4π = =. 4π The electic field due to a unifomly chaged sphee at a point outside it, is identical with the field due to an eual point chage placed at the cente. Page 7 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

8 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: Field at an intenal point: Suppose, we wish to find the electic field at a point P inside the spheical chage distibution. We daw a spheical suface passing though P and concentic with the given chage distibution. The adius of this sphee will be. All the points of this sphee ae euivalent. By symmety, the field is adial at P all the points of this suface and has a constant magnitude. The flux though this spheical suface is φ =. d S = ds = ds φ = 4...(1) π Let us now calculate the total chage contained inside this spheical suface. As the chage is unifomly distibuted within the given spheical volume, the chage pe unit volume is. The volume losed by the Gaussian 4 π 4 suface, though which the flux is π. He, the chage losed is 4 =. π =...(ii) 4 π Using Gauss s law, (i) & (ii), φ = 4π = = 4π The electic field due to a unifomly chaged sphee at an intenal point is in adial diection. 4π Vaiation of electic field with distance fom the cente is shown in following gaph: lectic Field due to a Plane Sheet of Chage Conside a lage plane sheet of chage with unifom suface chage density σ. We have to find the electic field at a point in font of the sheet. The cylinde togethe with its coss-sectional aeas foms a closed suface and we apply Gauss s law to this suface. A P The electic field at all the points on the ight face has the same magnitude. Thus, the flux of the electic field though ight suface is A. = A Similaly, the flux of the electic field though left face is also A. At the points on the cuved suface, the field and the outwad nomal make an angle of 9 o with each othe and he, flux A=. The total flux though the complete closed suface is φ =. d S = A A φ = A (i) The aea of the sheet losed in the cylinde is A. The chage contained in the cylinde is, theefoe, = σ A (ii) He fom Gauss s law, A φ = A = σ σ = We see that the field is unifom and does not depend on the distance fom the chage sheet. This is tue as long as the sheet is lage as compaed to its distance fom P. ds A Page 8 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

9 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: lectic Field nea a Chaged Conducting Suface Let us conside a lage, plane conducting sheet. The suface on ight has a unifom suface chage density σ. We have to find the electic field at a point P nea this suface and outside the conducto. As we know, the conducting suface is an euipotential suface and the electic field nea the suface is pependicula to the suface. Fo positive chage on the suface, the field is away fom the suface. To find the electic field, we constuct a Gaussian suface as shown in figue. We daw a cylinde whose left side is teminated in the inteio of the conducting sheet. The electic field at all the points on the ight face has the same magnitude. Thus, the flux of the electic field though ight suface is. A = A At the points on the cuved suface, the field and the outwad nomal make an angle of 9 o with each othe and he, flux A=. Also, the flux on the left face is zeo as the field inside the conducto is zeo in electostatic. The total flux though the Gaussian suface constucted is, theefoe, φ = A (i) The chage losed inside the closed suface is = σ A (ii) He fom Gauss s law, A φ = A = σ = σ The electic field nea a chaged conducting suface is σ and it is nomal to the suface. n A = P A uick ecap Aea vecto of a flat suface is a vecto pependicula to suface having magnitude eual to its aea A. Aea vecto is taken in outwad nomal diection fo a closed suface. lectic flux: The total numbe of electic field lines passing nomally though an aea placed inside an electic field is called electic flux. φ = A. (lectic flux though a flat aea in a unifom field) φ = (lectic flux fo geneal case) φ = (lectic flux though a closed suface) Gauss s Law: Gauss's law states that the total electic flux 1 though any closed suface is times total chage losed by that suface. Thus, φ = ( Gauss's Law ). = ( Gauss's Law ) Impotant points to note about Gauss s Law: 1. φ = electic flux because of electic field ceated by all chages (inside and outside the Gaussian suface). = total chages which ae inside the Gaussian suface. Gauss s Theoem fo any medium: φ = = Total flux Flux density = Aea λ lectic field due to a infinite long chage: = π lectic field due to a non-conducting infinite plane sheet σ of chage: = lectic field nea a conducting suface: = σ lectic Field due to a unifomly chaged Spheical Shell Inside: = ; Outside: =. 4π lectic Field due to a unifomly chaged non-conducting solid sphee Inside: = ; Outside: =. 4π 4π lectic Field due to a conducting solid sphee Inside: = ; Outside: =. 4π Page 9 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

10 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: xample Find the flux of the electic field though a spheical suface of adius due to a chage of 1-7 C at the cente and anothe eual chage at a point away fom the cente. Solution: Hee only one chage is inside the sphee. He, fom Gauss s Law, flux though sphee φ = = 7 1 φ = φ Nm C xample. Show that thee can be no net chage in a egion in which the field is unifom at all points. Solution: We know that the net flux though the suface is popotional to the net numbe of lines leaving the suface, whee the net numbe means the numbe leaving the suface minus the numbe enteing the suface. Unifom electic field means that electic field lines ae paallel and spacing between any two field line is same. He, if a closed suface is placed in a egion of unifom electic field, the net numbe of lines leaving the suface is zeo, since same numbe of field lines leave as that which ente. Thus, flux ø though a closed suface in a unifom electic field is zeo. He, fom Gauss s Law φ = = =. He, thee can be no net chage in a egion in which the field is unifom at all points. xample 4 A chage is placed at a distance a above the cente of a hoizontal, suae suface of edge a as shown in figue. Find the flux of the electic field though the suae suface. a / a Solution: This poblem is bit ticky. The key idea is that we can find flux using φ =. but that pocess will be uite cumbesome. Anothe key idea is that we can use Gauss s Law to find the flux; we need a closed suface fo that. One impotant concept is that we should use Gauss s Law fo the case when the situation is symmetical. So, what to do? As the suae suface has edge a and also the chage is kept at a distance a fom the cente of the suface, we can constuct a cube of edge a having one face coincident with given suface, such that it loses chage at the cente. He, fom Gauss s Law, flux though the entie cube φ = = As the situation is symmetical each face will shae same amount of flux. He, flux though euied suface Total flux = = 6 6 uick xecise 1. Fou closed sufaces, S 1 though S 4, togethe with the chages -,, and - ae sketched in Figue. Find the electic flux though each suface.. Figue shows six point chages that all lie in the same plane. Five Gaussian S 1 though S 5, each lose pat of this plane, and Fig. shows the intesection of each suface with the plane. ank these five sufaces in ode of the electic flux though them, fom most positive to most negative. a Page 1 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

11 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: xample 4. A chage is distibuted unifomly within the mateial of a hollow sphee of inne and oute adii 1 & espectively. Find the electic field at a point P a distance x away fom the cente fo 1 < x <. Daw a ough diagam showing the electic field as a function of x fo < x < Solution: This poblem is a simple application of Gauss s Law. Let us conside a Gaussian spheical suface at of adius x concentic with given sphee. To use Gauss s law we must have to find out the total chage inside the Gaussian suface. As the given sphee is hollow, its volume is given by : ( 1 ) 4π V = Theefoe, chage density = = V 4π ( 1 ) He, chage inside the Gaussian suface = (chage density) (volume of shell with inne and oute adii 1 & x) ( ) ( ) ( ) ( ) ( ) 4π = = 1 1 4π (1) He, using Gauss s Law, flux though Gaussian suface: ( x 1 ) φ = =...() ( 1 ) As the situation is symmetical, field at any point must be along adial diection. Since, aea vecto is in the outwad nomal diection, aea vecto ds at each point of Gaussian suface will be along. (Since magnitude of field is constant at each point of Gaussian Suface) Φ = S = (4ππ x ) () (suface aea of Gaussian suface = 4 ππ x ) He, using () & (), we get ( x 1 ) ( 4π x ) = = 4π ( ) ( x 1 ) x ( 1 ) 1...(4) Plotting the electic field as a function of x fo < x < Fo egion x < 1 : Sphee is hollow, theefoe thee can t be any chage inside it. He, thee can t be any electic field inside the egion x < 1. Plot is oughly shown (blue coloed) Fo egion 1 < x < : lectic field is given by expession (4) of the last slide. Thus, electic field incease hee. Plot is toughly shown (ed coloed). Fo egion x > : Outside egion of sphee. Hee, spheical shell chage behaves as if the net chage is placed at the cente. Thus field at a distance x fom cente outside the sphee is given by: =. 4π x Thus field deceases with distance which can be plotted oughly (sky blue). He, flux though Gaussian suface is φ =. d S = ds cos = ds (because angle between & ds = ) φ = ds Page 11 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

12 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: xecises NCT Poblems Caeful measuement of the electic field at the suface of a black box indicates that the net outwad flux though the suface of the box is 8 1 Nm C -1. (a) What is the net chage inside the box? (b) If the net outwad flux though the suface of the box wee zeo, could you conclude that thee wee no chages inside the box? Why o Why not? A point chage 1 μc is a distance 5 cm diectly above the cente of a suae of side 1 cm. What is the magnitude of the electic flux though the suae? A point chage of. μc is at the cente of a cubic Gaussian suface 9. cm on edge. What is the net electic flux though the suface? 1.. A point chage causes an electic flux of 1. 1 Nm C -1 to pass though a spheical Gaussian suface of 1. cm adius cented on the chage. (a) If the adius of the Gaussian suface wee doubled, how much flux would pass though the suface? (b) What is the value of the point chage? 1.1. A conducting sphee of adius 1 cm has an unknown chage. If the electic field cm fom the cente of the sphee is N/C and points adially inwad, what is the net chage on the sphee? 1.. A unifomly chaged conducting sphee of.4 m diamete has a suface chage density of 8. μc/m. (a) Find the chage on the sphee. (b) What is the total electic flux leaving the suface of the sphee? 1.. An infinite line chage poduces a field of N/C at a distance of cm. Calculate the linea chage density Two lage, thin metal plates ae paallel and close to each othe. On thei inne faces, the plates have suface chage densities of opposite signs and of magnitude 17 1 C/m. What is? (a) in the oute egion of the fist plate, (b) in the oute egion of the second plate, and (c) between the plates? 1.8. (a) A conducto A with a cavity as shown in Fig. 1.6(a) is given a chage. Show that the entie chage must appea on the oute suface of the conducto. (b) Anothe conducto B with chage is inseted into the cavity keeping B insulated fom A. Show that the total chage on the outside suface of A is [Fig. 1.6(b)]. (c) A sensitive instument is to be shielded fom the stong electostatic fields in its envionment. Suggest a possible way. Moe poblems fo pactice: 1. Conside Gauss law. ds =, then fo the situation shown in the figue at though Gaussian suface, (a) the due to would be zeo (b) the due to both 1 and would be non-zeo (c) the φ due to 1 and 1 would be on-zeo (d) the φ due to would be zeo. In a egion of the space, the electic field is in the x- diection and popotional to x, i. e., = xi. Conside an imaginay cubical volume of edge a, with its edges paallel to the axes of coodinates. The chage inside this volume is (a) zeo (b) a (c) a (d) 1 a 6. A chage is placed at the cente of the mouth of a conical flask. The flux of the electic field though the flask is (a) Zeo (b) (c) (d) < 1.9. A hollow chaged conducto has a tiny hole cut into its σ suface. Show that the electic field in the hole is = n, whee n is the unit vecto in the outwad nomal diection, and σ is the suface chage density nea the hole. 1.. Obtain the fomula fo the electic field due to a long thin wie of unifom linea chage density λ without using Gauss s law. 4. A long sting with a chage of λ pe unit length passes though an imaginay cube of edge a. The maximum flux of the electic field though the cube will be λa (a) (b) (c) (d) λa 6λa λa Page 1 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

13 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: Chages 1 and lie inside and outside espectively of a closed sufaces S. Let be the field at any point on S and φ be the flux of ove S. (a) If 1 changes, both and φ will change. (b) If changes, will change but φ will not change. (c) If 1 = and then but φ =. (d) If 1 and = then = but φ. 6. Chages 1 and ae placed inside and outside espectively of an unchanged conducting shell. Thei sepaation is. (a) The foce on 1 is zeo. (b) The foce on 1 is k k (c) The foce on is (d) The foce on is zeo A spheical conducto A lies inside a hollow spheical conducto B. Chages 1 and ae given to A and B espectively. (a) Chage 1 will appea on the oute suface of A. (b) Chage 1 will appea on the inne suface of B. (c) Chage will appea on the oute suface of B. (d) Chage 1 will appea on the oute suface of B. 8. A disk of adius a / 4 having a unifomly distibuted chage 6 C is placed in the x y plane with its cente at ( a /,, ). A od of length a caying a unifomly distibuted chage 8C is place on the x-axis fom x = a / 4 to x = 5a / 4. Two point chages 7 C and C ae placed at (a / 4, a / 4, ) and ( a/4, a / 4, ), espectively. Conside a cubical suface fomed by six sufaces x = ± a /, y = ± a /, z = ± a /. The electic flux though this cubical suface is (IIT-J 9) (a) C / (b) C / (c) 1 C / (d) 1 C / 9. A cubical egion of side a has its cente at the oigin. It loses thee fixed point chages, at (, a/4, ), at (,, ) and at (, a/4, ). Choose the coect option(s). (IIT-J 1) a) The net electic flux cossing the plane x = a/ is eual to the net electic flux cossing the plane x = a/. b) The net electic flux cossing the plane y = a/ is moe than the net electic flux cossing the plane y = a/. c) The net electic flux cossing the entie egion is d) The net electic flux cossing the plane z = a/ is eual to the net electic flux cossing the plane x = a/. 1. A chage is placed at the cone of a cube. The flux of the electic field though the cube is (a) Zeo (b) (c) (d) 8 Page 1 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao

14 Gauss s Law Autho: P. K. Bhati (IIT Khaagpu) Mb: ADMISSION NOTIC Fo XIIth moving students Physics & Mathematics By Panjal Si (B. Tech., I.I.T. Khaagpu) WKend Batch fom 1 th Ap nd egula Batch fom 4 th Ap Fee stuctue (Boad J Main J Advanced) Batch: s. 15 (Boad J Main) Batch: s. 1 Boad Level Batch: s. 1 Weekend Batch: s. 1 Concession in Fee fo economically weak students. ADMISSION NOTIC Fo XIth moving students School Coaching Pogam fom th Ap We have launched a new concept to povide Schooling & Coaching at same place. It will save time and money of students. It will also minimize stess level of students. It is diffeent fom syncho pogam. We will cove Boad level as well as competition level. We will stat fom basics and gadually maste J Advanced level poblems. Fee stuctue: Admission Fee: s. Secuity deposit: s. 1 Monthly Fee: s. pe month Total: s 4 = s.7 eadmission fee in 1 th : s. 15 Miscellaneous chages: s. 8 Total fo yeas: s. 15 Mb: mail: pkbhati.iit@gmail.com Website: Addess: Concept, JB, Top Floo, Nea Jitenda Cinema, Secto 4, Bokao Steel City Mb: mail: pkbhati.iit@gmail.com Website: Addess: Concept, JB, Top Floo, Nea Jitenda Cinema, Secto 4, Bokao Steel City Page 14 CONCPT: JA-8, 1 st Floo, Behind Devi Cinema, City Cente, Bokao