Chapter # 30
Gauss’s Law
SOLVED EXAMPLES Example 30.1 A square frame of edge 10 cm is placed with its positive normal making an angle of 600 with a uniform electric field of 20 V/m. Find the flux of the electric field through the surface bounded by the frame. Sol. The surface considered is plane and the electric field is uniform figure. Hence, the flux is = E . S = E S cos 600 1 = (20 V/m) (0.01 m 2 ) = 0.1 V–m. 2
normal field 600
10 cm
10 cm Example 30.2 A change q is placed at the centre of a sphere. Taking outward normal as positive, find the flux of the electric field through the surface of the sphere due to the enclosed charge. Sol.
E
s
r q
Let us take a small element S on the surface of the sphere figure. The electiric field here is radially outward and has the magnitude q 4 0 r 2
,
where r is the radius of the sphere. As the positive normal is also outward, = 0 and the flux through this part is q S. = E. S = 4 0 r 2
Summing over all the parts of the spherical surface, =
q 4 0 r
2
S
q
=
4 0 r
2
q 4r 2 = . 0
Example 30.3 A uniform electric field exists in space. Find the flux of this field through a cylindrical surface with the axis parallel to the field. Sol.
Normal S E manishkumarphysics.in
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Chapter # 30
Gauss’s Law
Consider figure and take a small area S on the cylindrical surface. The normal to this area will be perpendicullar to the axis of the cylinder. But the electric field is parallel to the axis and hence = E. S = E S cos ( / 2) = 0. This is true for each small part of the cylindrical surface. Summing over the entire surface, the total flux is zero. Example 30.4 A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is constructed with its centre at the periphery of the ring figure. Find the flux of the electric field through the surface of the sphere.
Ring
A 0 01 B
Sphere
Sol. From the geometry of the figure, OA = OO1 and O 1 A = O1 O Thus, OAO1 is an equilateral triangle.Hence AOO1 = 600 or AOB = 1200. The are AO1 B of the ring subtends an angle 1200 at the centre O. Thus, one third of the ring is inside the sphere. The charge enclosed by the sphere =
Q . From Gauss’s law, the flux of the electrical field through the 3
Q surface of the sphere is 3 . 0
WORK OUT EXAMPLES 1.
A uniform electric field of magnitude E = 100 N/C exist in the space in X-direction. Calculate the flux of this field through a plane square area of edge 10cm placed in the Y–Z plane. Take the normal along the positive X-axis to be positive
Sol.
The flux = E cos dS. As the normal to the area points along the electric field, = 0. Also E is uniform, so = ES 2 = (100 N/C) (0.10 m)2 = 1.0 N m . C
2.
Sol.
A large plane charge sheet having having surface charge density = 2.0 × 10–6/Cm 2 lies in the X-Y plane. Find the flux of the electric field through a circular area of radius 1 cm lying completely in the region where x, y, z are all positive and with its normal making an angle of 60º with the Z-axis. The electric field near the plane charge sheet is E = /2 0 in the direction away from the sheet. At the given area, the field is along the Z-axis. The area = r2 = 3.14 × 1cm 2 = 3.14 × 10–4ms 2. The angle between the normal to the area and the field is 60º
Hence, the flux = E . S = ES cos = e r2 cos 60º 0 =
2.0 10 6 C / m 2 12
2 8.85 10 C / N m = 17.5 N–m 2/C.
3.
2
2
× (3.14 × 10–4 m 2)
1 2
A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of radius 5cm. (a) Find the electric field at a point 2 cm away from the centre. (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere. manishkumarphysics.in
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Chapter # 30
Gauss’s Law
(a) Let us consider figure. Suppose, we have to find the field at the point P. Draw a concentric spherical surface through P. All the points on this surface are equvalent and by symmetry, the field at all these points will be equal in magnitude and redial in direction. The flux through this surface = E.dS
=
E.dS
= E dS
= 4x 2 E, where x = 2cm = 2 × 10–2 m. From Gauss’s law, this flux is equal to the charge q contained inside the surface divided by 0. Thus, 4x 2 E = q/ 0 or,
E
q 4 0 x 2
2 4 10 8 C 9 Nm = 9 10 C 2 4 10 4 m2
= 9 × 105 N/C. (b) See figure. Take a Gaussian surface through the material of the hollow sphere. As the electric field in a conducting material is zero, the flux E.dS through this Gaussian surface is zero. Using Gauss’s
law, the total charge enclosed must be zero. Hence, the charge on the inner surface of the hollow sphere is –4 × 10–8 C. But the total charge given to this hollow sphere is 6 × 10–8 C.Hence, the charge on the outer surface will be 10 × 10–8 C. 4.
Figure shows three concentric thin spherical shells A, B and C of radii a, b and c respectively. The shells A and C are given charges q and – q respectively and the shell B is earthed. Find the charged appearing on the surfaces of B and C.
Sol.
As shown in the previous worked out example, the inner surface of B must have a charge –q from the Gauss’s law. Suppose, the outer surface of B has a charge q’. The inner surface of C must have a charge –q’ from the Gauss’s law. As the net charge on C must be –q, its outer surface sould have a charge q’ – q. The charge distribution is shown in figure. The potential at B due to the charge q on A
q = 4 b , 0 due to the charge –q on the inner surface of B
q = 4 b , 0 due to the charge –q’, on the inner surface of C manishkumarphysics.in
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Chapter # 30
Gauss’s Law
q' = 4 c , 0 and due to the charge q’ – q on the outer surface of C
q q = 4 c 0 The net potential is
VB
q q 4 0b 4 0 c .
This should be zero as the shell B is earthed. Thus, b q. c The charge on various surfaces are as shown in figure q
5.
An electric dipole consists of charges ±2.0 × 10 8C separated by a distance of 2.0 × 10–3m. It is placed near a long line charge of linear charge density 4.0 × 10–4 C/m as shown in figure, such that the negative charge is at a distance of 2.0 cm from the line charge. Find the force acting on the dipole.
Sol.
The electric field at a distance r from the line charge of linear density is given by
E
2 0 r
Hence, the field at the negative charge is
( 4.0 10 4 C / m)(2 9 10 9 N m 3 / C 2 ) 0.02m The force on the negative charge is F1 = (3.6 × 108 N/C) (2.0 × 10–8 C) = 7.2 N towards the line charge. Similarly, the field at the positive charge, i.e., at r = 0.022 m is E2 = 3.3 × 108 N/C. The force on the positive charge is F2 = (3.3 × 108 N/C) (2.0 × 10–8 C) = 6.6 M away from the line charge. Hence, the net force on the dipole = (7.2 – 6.6) N = 0.6 N towards the line charge. E1
6. Sol.
The electric field in a region is radially outward with magnitude E = Ar. Find the charge contained in a sphere of radius a centred at the origin. Take A = 100 V/m 2 and a = 20.0 cm. The electric field at the surface of the sphere is Aa and being radial it is along the outward normal. flux of the electric field is, therefore,
=
E dS
cos = Aa (4 a2).
The charge contained in the sphere is, from Gauss’s law, Qinside = 0 = 4 0 Aa3
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Chapter # 30
Gauss’s Law
1 C2 9 10 9 N m 2 = 8.89 × 10–11 C.
100 V (0.20 m)2 m2
7.
A particle of mass 5 × 10–6 g is kept over a large horizontal sheet of charge of density 4.0 × 10–6 C/m 2 figure. What charge should be givento this particle so that if released, it does not fall down? How many electrons are to be removed to give this charge? How much mass is decreased due to the removal of these electrons?
Sol.
The electric field in front of the sheet is
E
4.0 10 6 C / m2 2 0 2 8.85 10 12 C2 / N m2
= 2.26 × 105 N/C. If a charge q is given to the particle, the electric force qE acts in the upward direction. It will balance the weight of the particle if q × 2.26 × 105 N/C = 5 × 10–9 kg × 9.8 m/s 2 q
4.9 10 8
C 2.26 10 5 = 2.21 × 10–13 c. The charge on one electron is 1.6 × 10–19 C. The number of electrons to be removed
or,
2.21 10 13 C
= 1.4 × 106. 1.6 10 19 C Mass decreased due to the removal of these electrons = 1.4 × 106 × 9.1 × 10–31 kg = 1.3 × 10–24 kg. =
8. Sol.
Two conducting plates A and B are placed parallel to each other. A is given a charge Q 1 and B a charge Q2. Find the distribution of charges on the four surfaces. Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss’s law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B.
The distribution should be like the one shown in figure. To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation E = / (2 0), the electric field at P
Q1 q due to the charge Q 1 – q = 2 A (downward) 0 q due to the charge + q = 2A (upward), 0 q due to the charge – q = 2A (downward), 0 Q2 q and due to the charge Q 2 + q = 2A n (upward). 0 The net electric field at P due to all the four charged surfaces is (in the downward direction) manishkumarphysics.in
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Chapter # 30
Gauss’s Law
Q1 q Q q q q 2 2 A 0 2 A 0 2 A 0 2 A 0 As the point P is inside the conductor, this field should be zero. Hence, Q1 – q – Q2 – q – 0
q
or, Thus,
Q1 q
Q1 Q 2 2
Q1 Q 2 2
..............(i)
Q1 Q 2 ..............(ii) 2 using these equation, the distribution dhown in the figure can be redrawn as in figure and
Q2 q
This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost, surfaces get equal charges and the facing surfaces get equal and opposite charges.
QUESTIONS FOR SORT ANSWERS 1.
A small plane area is rotated in an electric field. In which orientation of the area is the flux of electric field through the area maximum? In which orientation is it zero?
2.
A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through 180º . Does the flux of electric field change? If yes, does it decrease or increase?
3.
A charge Q is uniformly distributed on a thin spherical shell. What is the field at the centre of the shell? If a point charge is brought close to the shell, will the field at the centre change? Does your answer depend on whether the shell is conducting or nonconducting?
4.
A spherical shell made of plastic, contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it without altering the charge, will the field inside be changed? What happens if the shell is made of a metal?
5.
A point charge q is placed in a cavity in a metal block. If a charge Q is brought outside the metal, will the charge q feel an electric force?
6.
A rubber balloon is given a charge Q distributed uniformly over its surfaces. Isthe field inside the balloon zero everywhere if the balloon does not have a spherical surface?
7.
It is said that any charge given to a conductor comes to its surfaces. Should all the protons come to the surface? Should all the electrons come to the surface? Should all the free electrons come to the surface?
OBJECTIVE I 1.
A charge Q is unifromly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become IykfLVd dh ,d cM+h IysV ij vkos'k Q leku :i esa forfjr gSA IysV ds dsUnz ds fudVLFk fcUnq ij fo|qr {ks=k 10 V/m gSA ;fn IykfLVd dh txg leku T;kferh; foekvksa dh ,d rkacs dh IysV yh tk,] ftl ij vkos'k Q gh gks rks P ij fo|qr {ks=k dk eku gks tk;sxk (A) zero (B) 5 V/m (C*) 10 V/m (D) 20 V/m manishkumarphysics.in
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Chapter # 30 2.
Gauss’s Law
A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be
,d /kkfRod d.k] ftl ij dksbZ ifj.kkeh vkos'k ugha gS] ,d /ku vkosf'kr ifjfer /kkrq dh IysV dh lehi j[kk tkrk gSA d.k ij oS|qr cy gksxk (A*) towards the plate (B) away from the plate (C) parallel to the plate (D) zero (A*) IysV dh rjQ (B) IysV ds lekUrj (C) IysV ds lekUrj (D) 'kwU; 3.
A thin, metallic spherical shell contains a chrage Q on it. A point charge q is placed at the centre of the shell and another charge q1 is placed outside it as shown in fig. All the three charges are positive. The force on the charge at the centre is ,d irys /kkfRod xksykdkj dks'k ij Q vkos'k gSA ,d fcUnq vkos'k q dks'k ds dsUnz ij j[kk tkrk gS rFkk nwljk vkos'k q1 blds ckgj fp=kkuqlkj j[kk tkrk gSA lHkh rhuksa vkos'k /kukRed gSA dsUnz ij fLFkr vkos'k ij oS|qr cy gS -
(A) towards left (A) cka;h rjQ
(B) towards right (B) nka;h rjQ
(C) upward (C) Åij dh
rjQ
(D*) zero (D*) 'kwU;
4.
Consider the situation of the previous problem. The force on the centre charge due to the shell is (A) towards left (B*) towards right (C) upward (D) zero iwoZ iz'u esa dsUnzh; vkos'k ij dks'k ds dkj.k yxus okyk cy gS (A) cka;h rjQ (B*) nka;h rjQ (C) Åij dh rjQ (D) 'kwU;
5.
Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20cm will be ,d NksVs ls vk;ru esa fo|qr vkos'k forfjr gSA dqy vkos'k dks ifjc) djus okys 10lseh f=kT;k ds ,d xksyh; i`"B ls fuxZr oS|qr ¶yDl 25 V-m gSA 20 lseh f=kT;k ds ,d ladsUnzh xksys ls fuxZr ¶yDl gksxk (A*) 25 V-m (B) 50 V-m (C) 100 V-m (D) 200 V-m
6.
Fig. shows an imaginary cube of edge L/2. A unifromly charged rod of length L moves towards left at a small but constant speed v. At t = 0, the left end just touches the centre ot the face of the cube opposite it. Which of the graphs shown in fig. represents the flux of the electric filed through the cube at the rod goes through it ? ) iznf'kZr fp=k (a) L/2 Hkqtk ds ,d dkYifud ?ku dks n'kkZrk gSA L yEckbZ dh ,d leku :i ls vkosf'kr ,d NM+ cka;h rjQ /khes ysfdu fu;r osx v ls xfr djrh gSA t = 0, ij NM+ dk cka;k fljk ?ku ds blds lkeus okys Qyd ds dsUnz dks Bhd Li'kZ djrk gSA tSls&tSls NM+ blesals xqtjrh gS] rks fp=k (b) esa n'kkZ, ys[kkfp=kksa esals dkSulk ?ku fuxZr oS|qr ¶yDl dks n'kkZrk gS -
(a)
(b)
(Ans : d 7.
A charge q is placed at the centre of the open end of a cylindrical vessel as shown in fig. The flux of the electric field through the surface of the vessel is HCV_Ch-30_Obj.I_7 fp=k es n'kkZ, vuqlkj csyukdkj ik=k ds [kqys fljs ij ,d vkos'k q j[kk gqvk gSA ik=k ds lrg ls ikfjr fo|qr {ks=k dk ¶yDl
gksxk &
(A) zero 'kwU;
(B) q/ 0
(C*) q/20
manishkumarphysics.in
(D) 2q/0 Page # 7
Chapter # 30
Gauss’s Law
OBJECTIVE II 1. Mark the correct options : lgh fodYi pqfu;s (A) Gausss’s law is valid only for symmetrical charge distributions (B) Gauss’s law is valid only for charges placed in vacuum (C) The electric field calculated by Gauss’s law is the field due to the charges inside the Gaussian surface (D*) The flux of the electric field through a closed surface due to all the charge is equal to the flux due to the charges enclosed by the surface. (A) xkÅl dk fu;e dsoy lefer vkos'k forj.kksa ds fy, oS/k gSA (B) xkÅl dk fu;e dsoy fuokZr esa j[ks vkos'kksa ds fy, oS/k gSA (C) xkÅl ds fu;e ls laxf.kr fo|qr {ks=k] xkÅlh; i`"B ds Hkhrjh vkos'kksa ds dkj.k mRiUu {ks=k gSA (D*) ,d can i`"B ls lHkh vkos'kksa ds dkj.k fuxZr oS|r q ¶yDl i`"B ds }kjk ifjc) vkos'kksa ds dkj.k ¶yDl ds cjkcj
gSA
2. A positive point charge Q is brought near an isolated metal cube. (A) The cube becomes negatively charged (B) The cube becomes positively charged (C) The interior becomes positively charged and the surface becomes negatively charged. (D*) The interior becomes remains charged free and the surface gets nonuniform charge distribution. ,d /kukRed fcUnq vkos'k Q ,d foyfxr /kkfRod ?ku ds fudV yk;k tkrk gS] rks (A) ?ku _.kkosf'kr gks tkrk gSA (B) ?ku /kukosf'kr gks tkrk gSA (C) vkarfjd Hkkx /kukosf'kr rFkk i`"B _.kkosf'kr gks tkrk gSA (D*) vkarfjd Hkkx vkos'k eqDr jgrk gS rFkk i`"B ij vleku forj.k gks tkrk gSA 3. A large nonconducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in fig. ,d cM+h vpkyd ijr dks ,dvleku :i ls vkos'k /kuRo fn;k tkrk gSA nks vukosf'kr /kkrq dh NksVh NM+as A o B fp=kkuqlkj ijr ds utnhd j[kh tkrh gS -
(A*) M attracts A (B*) M attracts B (C*) A attracts B (A*) M, A dks vkdf"kZr djrh gSA (B*) M, B dks vkdf"kZr djrh (C*) A attracts B (D*) B, A dks vkdf"kZr djrh
(D*) B attracts A
gS gS
4. If flux of the electric field through a closed surface is zero, (A) the electric field must be zero everywhere on the surface (B*) the electric field may be zero everywhere in the surface (C*) the charge inside the surface must be zero (D) the charge in the vicinity of the surface must be zero. ;fn ,d can i`"B ls fuxZr oS|qr ¶yDl 'kwU; gS] rks (A) i`"B ij gj txg fo|qr {ks=k 'kwU; gksuk pkfg,A (B*) i`"B ij gj txg fo|qr {ks=k 'kwU; gks ldrk gSA (C*) i`"B ds Hkhrj vkos'k 'kwU; gksuk pkfg,A (D) i`"B ds bnZ&fxnZ vkos'k 'kwU; gksu k pkfg,A 5. An electric dipole is placed at the cnetre of a sphere, Mark the correct options. (A*) The flux of the electric filed through the sphere is zero (B) The electric field is zero at every point of the sphere (C*) The electic field is not zero anywhere on the sphere (D) The electric field is zero on a circle on the sphere. ,d oS|qr f}/kzqo ,d xksys ds dsUnz ij j[kk gS] lgh fodYi pqfu;s -
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Chapter # 30
Gauss’s Law
(A*) xksy s ls fuxZr oS|qr ¶yDl 'kwU; gSA (B) xksys ds izR;sd fcUnq ij fo|qr {ks=k 'kwU; gSA (C*) xksys ij dgha Hkh fo|qr {ks=k 'kwU; ugha gSA (D) xksys ij fLFkr ,d o`Ùk ij oS|qr {ks=k 'kwU; gSA 6. Figure (30-Q5) shows a charge q placed at the centre of a hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged ? ,d vkos'k q fp=kkuqlkj ,d xksyk/kZ ds dsUnz ij j[kk x;k gSA ,d nwljk vkos'k Q fLFkfr;ksa A, B, C o D esals ,d ij j[kk gSA nwljs vkos'k dh fdl fLFkfr@fLFkfr;ksa esa xksyk/kZ ls fuxZr ¶yDl vifjofrZr jgsxk -
(A*) A
(B) B
(C*) C
(D) D
7. A closed surface S is constructed around a conducting wire connected to a battery and a switch (fig.) As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electorns leaving it. On closing the switch, the flux of the electric field through the closed surface.
,d cSVjh ,oa ,d fLop ls tqM+s ,d pkyd rkj ds ifjr% ,d can i`"B dh jpuk ¼fp=kkuqlkj½ dh tkrh gSA tSls gh fLop can fd;k tkrk gSA rkj esa eqDr bysDVªkWu rkj ds vuqfn'k xfr djuk izkjEHk dj nsrs gSaA fdlh Hkh le;kUrjky esa can i`"B S esa izos'k djus okys bysDVªkWuksadh la[;k blls ckgj fudyus okys bysDVªkWuksa dh la[;k blls ckgj fudyus okys bysDVªkWuksa dh la[;k ds cjkcj gksrh gSA fLop can djus ij can i`"B ls ikfjr oS|qr ¶yDl -
(A) is increased (A) c<+rk gS
(B) is decreased (B) ?kVrk gS
(C*) remains unchanged (C*) vifjofrZr jgrk gSA
(D*) remains zero (D*) 'kwU; jgrk gSA
8. Fig. shows a closed surface which intersects a conducting sphere. If a positive charged is placed at the point P, the flux of the electic field through the closed surface fp=k esa ,d can i`"B n'kkZ;k x;k gS tks ,d pkyd xksys dks izfrPNsn djrk gSA ;fn fcUnq P ij ,d /kukRed vkos'k j[kk tk, rks can i`"B ls ikfjr oS|qr ¶yDl —
(A) will remain zero (B*) will become positive (A) 'kwU; jgsxk (B*) /kukRed gks tk,xk
(C) will become neagative (D) will become undefined (C) _.kkRed gks tk,xk (D) vifjHkkf"kr gksxk
EXERCISE 1.
The electric field in a region is given by E 3 E 0 i 4 E 0 j with E0 = 2.0 × 103 N/C. Find the flux of this 5 5
field through a rectangular surface of area 0.2m 2 parallel to the Y–Z plane.
3 5
4 5
[Ans. 240
N m2 ] C
,d {ks=k esa fo|qr {ks=k dk eku E E 0 i E 0 j }kjk fn;k tkrk gS] tgk¡ E0 = 2.0 × 103 N/C gSA blds Y-Z ry ds lekUrj 0.2m 2 ds ,d vk;rkdkj i`"B ls fuxZr oS|qr ¶yDl Kkr dhft,A manishkumarphysics.in
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Chapter # 30 2.
Gauss’s Law
A charge Q is uniformy distributed over a rod of length . Consider a hypothetical cube of edge with the centre of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube. [Ans. Q/(e 0)] ,d vkos'k Q, yEckbZ dh ,d NM+ ij leku :i ls forfjr gSA Hkqtk ds ,d dkYifud ?ku ij fopkj dhft,] ftldk
dsUnz NM+ ds ,d fljs ij gSA ?ku ds lEiw.kZ i`"B ls fuxZr U;wure laHko oS|qr ¶yDl Kkr dhft,A [Ans. Q/(e 0)] 3.
Show that there can be no net charge in a region in which the electric field is uniform at all points.
fn[kkb;s fd ,d {ks=k esa] ftlesa fo|qr {ks=k lHkh fcUnqvksa ij leku gks] dksbZ ifj.kkeh vkos'k ugha gks ldrkA 4.
E x 0 i . Find the charge contained inside a cubical volume The electric field in a region is given by E bounded by the surface x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103 N/C, = 2 cm and a = 1 cm. [Ans. 2.2 × 10–12 C]
,d {ks=k esa fo|qr {ks=k E
E0 x i
}kjk fn;k tkrk gSA x = 0, x = a, y = 0, y = a, z = 0 rFkk z = a i`"Bkksa }kjk ifjc)
,d ?kuh; vk;ru ds Hkhrj vkos'k Kkr dhft,A E0 = 5 × 103 N/C, = 2 cm rFkk a = 1 cm. [Ans. 2.2 × 10–12 C] 5.
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube. [Ans. Q/ 0] ,d vkos'k Q ,d ?ku ds dsUnz ij j[kk gSA ?ku ds N% i`"Bksa ls fuxZr oS|qr ¶yDl Kkr dhft,A [Ans. Q/ 0]
6.
A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in figure. Find the flux of the electric field through the square surface. [Ans. Q/(6 0)] ,d vkos'k Q fp=k esa n'kkZ, vuqlkj a Hkqtk ds ,d {kSfrt o.kkZdkj {ks=kQy ds dsUnz ls a/2 nwjh ij j[kk gSA oxkZdkj {ks=kQy ls ikfjr oS|qr ¶yDl Kkr dhft,A [Ans. Q/(60)]
7.
Find the flux of the electric field through a spherical surface of radius R due to a charge of 10–7C at the centre and another equal charge at a point 2R away from the centre dsUnz ij j[ks 10–7C ds vkos'k ,oa dsUnz ls 2R nwjh fLFkr fcUnq ij j[ks nwljs leku vkos'k ds dkj.k R f=kT;k ds xksyh; i`"B ls ikfjr oS|qr ¶yDl Kkr dhft, -
N m2 ] C A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss’s law, find the flux of the electric field due to this charge through the surface of the hemisphere ,d vkos'k Q ,d dkYifud v)Z xksyh; i`"B ds dsUnz ij j[kk gSA xkml dk fu;e ,oa T;kfefr rdZ iz;ksx djrs gq,] [Ans. 1.1 × 104
8.
bl vkos'k ds dkj.k v)Zxksyh; i`"B ls ifjr% oS|qr ¶yDl Kkr dhft,A
[Ans. Q/(2 0)] 9.
A spherical volume contains a uniformly distributed charge of density 2.0 × 10–4 C/m 3. Find the electric field at a point inside the volume at a distance 4.0 cm from the centre. ,d xksyh; vk;ru esa 2.0 × 10–4 C/m3 ?kuRo dk vkos'k leku :i ls forfjr gSA vk;ru ds Hkhrj dsUnz ls 4.0 lseh
nwj fLFkr fcUnq ij fo|qr {ks=k Kkr dhft,A
[Ans. 3.0 × 105 N/C] manishkumarphysics.in
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Chapter # 30 10.
Gauss’s Law
The radius of a gold nucleus (Z = 79) is about 7.0 × 10–15 m. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?[Ans. (a) 2.32 × 1021 N/C (b) 1.16 × 1021 N/C] lksus ds ukfHkd (Z = 79) dh f=kT;k yxHkx 7.0 × 10–15 m gSA eku yhft, fd /ku vkos'k laiw.kZ ukfHkdh; vk;ru esa
leku :i ls forfjr gSA fo|qr {ks=k dh rhozrk Kkr dhft, %
[Ans. (a) 2.32 × 1021 N/C (b) 1.16 × 1021 N/C] 11.
A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r 1 and r2 figure. Find the electric field at a point P a distance x away from the centre for r 1 < x < r2. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r 2. [Ans.
Q( x 3 r13 ) 4 0 x 2 (r23 r13 )
]
,d vkos'k Q, r1 vkarfjd o r2 cká f=kT;kvksa ds ,d [kks[kys xksys ds inkFkZ esa leku :i ls forfjr gSA dsUnz ls x nwjh ij fLFkr fcUnq P ij fo|qr {ks=k Kkr dhft,] tgk¡ r1 < x < r2 gSA oS|qr {ks=k ,oa x (tgk¡ 0 < x < 2r2) ds e/; ,d xzkQ [khafp,A
12.
[Ans.
Q( x 3 r13 ) 4 0 x 2 (r23 r13 )
A charge Q is placed at the centre of an uncharged, hollow metallic sphere of radius a (a) Find the surface charge density on the inner surface and on the outer surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and the outer surfaces? (c) Find the electric field inside the sphere at a distance x from the centre in the situations (a) and (b). ,d vkos'k Q, a f=kT;k ds ,d vukosf'kr [kks[kys xksys ds dsUnz ij j[kk gSA (a) Hkhrjh i`"B o cká i`"B ij i`"B vkos'k ?kuRo Kkr dhft,A (b) ;fn ,d vkos'k q xksys ij j[kk tk, rks Hkhrjh o cká i`"B ij i`"B vkos'k ?kuRo fdruk gksxk\ (c) fLFkfr;ksa (a) o (b) esa xksy s ds Hkhrj dsUnz ls x nwjh ij fo|qr {ks=k Kkr dhft,A [Ans. (a)
13.
14.
15.
]
Q
,
Q
, (b)
Q
,
Qq
Q
in both situations] 4 0 x 2 4a 4a 4a 4a Consider the following very rough modle of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius 10–15m. The two 1s electrons make a spherical charge cloud at an average distance of 1.3 × 10–11 m from the necleus, whereas the two 2s electrons make another spherical cloud at an average distance of 5.2 × 10–11 m from the nucleus. Find the electric field at (a) a point just inside the 1 s cloud and (b) a point just inside the 2s cloud. ,d csjhfy;e ijek.kq ds vR;ar jQ ekWMy ij fopkj dhft,A ukfHkd esa pkj izksVkWu ,oa pkj U;wVªkWu gS] tks 10–15 m f=kT;k ds ,d NksVs ls vk;ru esa lhfer gSA nks 1s bysDVªkWu ukfHkd ls 1.3 × 10–11 m dh vkSlr nwjh ij ,d xksyh; vkos'k vHkz cukrs gS tcfd nks 2s bysDVªkWu ukfHkd ls 5.2 × 10–11 m dh vkSlr nwjh ij nwljk xksyh; vkos'k vHkz cukrs gSaA fo|qr {ks=k Kkr dhft, : (a) 1 s vHkz ds Bhd vUnj ,d fcUnq ij (b) 2s vHkz ds Bhd vUnj ,d fcUnq ijA [Ans. (a) 3.4 × 1013 N/C, (b) 1.1 × 1012 N/C] 2
2
2
2
, (c)
Find the magnitude of the electric field at a point 4cm away from a line charge of density 2 × 10 –6 C/m. 2 × 10–6 c/m ?kuRo ds ,d js[kh; vkos'k ls 4cm nwj fcUnq ij fo|qr {ks=k dk ifjek.k Kkr dhft,A [Ans. 9 × 105 N/C] A long cylindrical wire carries a positive charge of linear density 2.0 × 10–8 C/m. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius. [Ans. 2.88 × 10–17 J] ,d yEcs csyukdkj rkj ij 2.0 × 10–8 c/m js[kh; ?kuRo dk ?kukRed vkos'k gSA ,d bysDVªkWu blds pkjksa vksj fLFkj
fo|qr vkd"kZ.k cy ds izHkko esa] ,d o`Ùkkdkj iFk ij pDdj yxkrk gSA bysDVªkWu dh xfrt ÅtkZ Kkr dhft,A /;ku nhft, fd ;g f=kT;k ij fuHkZj ugha djrh gSA [Ans. 2.88 × 10–17 J] 16.
A long cylindrical volume contains a uniformly distributed charge of density . Find the electric field at a point P inside the cylindrical volume at a distance x from its axis figure. [Ans. x/(2 0)] manishkumarphysics.in
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Chapter # 30
Gauss’s Law
,d yEcs csyukdkj vk;ru esa ?kuRo dk leku vkos'k forfjr gSA csyukdkj vk;ru ds Hkhrj blds v{k ls x nwjh ij fLFkr fcUnq P ij fo|qr {ks=k Kkr dhft,A [Ans. x/(2 0)]
17.
A nonconducting sheet of large surface area and thickness d contains uniform charge distribution of density . Find the electric field at a point P inside the plate, at a distance x from the central plane. Draw a qualitative graph of E against x for 0 < x < d. [Ans. x/ 0] ,d cM+s i`"Bh; {ks=kQy ,oa d eksVkbZ dh ,d vpkyd 'khV ij ?kuRo dk leku vkos'k forj.k gSA IysV ds Hkhrj dsUnzh; ry ls x nwjh ij fLFkr fcUnq P ij fo|qr {ks=k Kkr dhft,A E dk x ds lkFk xq.kkRed xzkQ [khafp,A [Ans. x/ 0]
18.
A charged particle having a charge of –2.0 × 10–6 C is placed close to a nonconducting plate having a surface charge density 4.0 × 10–6 C/m 2. Find the force of attraction between the particle and the plate. [Ans. ] –6 –6 2 –2.0 × 10 dwyke vkos'k dk ,d vkosf'kr d.k 4.0 × 10 dwyke/eh i`"B vkos'k ?kuRo okyh ,d dqpkyd IysV ds lehi j[kk gSA IysV o d.k ds e/; vkd"kZ.k cy Kkr dhft,A [Ans. ]
19.
One end of a 10cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate and the other end is fastened to a small ball having a mass of 10g and a charge of 4.0×10–6 C. In equilibrium, the thread makes an angle of 60º with the vertical. Find the surface charge density on the plate. [Ans. 7.5 × 10–7 C/m2] 10 lseh yEcs flYd ds /kkxs dk ,d fljk ,d vkosf'kr dqpkyd IysV ds ,d cM+s Å/okZ/kj i`"B ls tqM+k gSA nwljk fljk 10xzke nzO;eku o 4.0×10–6 dwykWe vkos'k dh ,d NksVh xsan ls tksM+k x;k gSA lkE;koLFkk esa /kkxk Å/okZ/kj esa 60º dk dks.k cukrk gSA IysV ij i`"B vkos'k ?kuRo Kkr dhft,A [Ans. 7.5 × 10–7 C/m2]
20.
Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations. iwoZ iz'u dh fLFkfr ij fopkj dhft, : (a) lkE;koLFkk esa Mksjh esa ruko Kkr dhft,A (b) eku yhft, fd xsan dks FkksM+k
lk ,d rjQ ys tkdj NksM+ fn;k tkrk gSA NksVs nksyuksa dk vkorZdky Kkr dhft,A
[Ans. (a) 0.20 N (b) 0.45 s] 21.
Two large conducting plates are placed parallel to each other with a separation of 2.00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces. [Ans. 0.505 × 10–12 C/m2] nks cM+h pkyd IysVsa ,d nwljs ds lekUrj ,d nwljs ls 2.00 lseh nwj j[kh gSA ,d bysDVªkWu ,d IysV ds ikl ls fojkekoLFkk ls izkjEHk dj nwljh IysV ij 2.00 ekbØks lsd.M esa igqaprk gSA vkarfjd i`"Bksa ij i`"B vkos'k ?kuRo Kkr dhft,A [Ans. 0.505 × 10–12 C/m2]
22.
Two large conducting plates are placed parallel to each other and they carry equal and opposite charges with surface density as shown in figure. Find the electric field (a) at the left of the plates , (b) in between the plates and (c) at the right of the plates. [Ans. (a) zero (b) / 0 (c) zero] nks cM+h pkyd IysVsa ,d nwljs ds lekUrj j[kh gS rFkk mu ij fp=kkuqlkj cjkcj ,oa foijhr vkos'k ?kuRo gSA fo|qr {ks=k Kkr dhft,A (a) IysVksa ds cka;h vksj (b) IysVksa ds e/; rFkk (c) IysVksa ds nka;h vksj [Ans. (a) zero (b) / 0 (c) zero]
23.
Two conducting plates X and Y, each having large surface area A (on one side), are placed parallel to each other as shown in figure. The plate X is given a charge Q whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plate X, (b) the electric field at a point to the left of the manishkumarphysics.in
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Chapter # 30
Gauss’s Law
plates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates. nks pkyd IysVsa X o Y ftudk izR;sd dk i`"B {ks=kQy ¼,d rjQ dk½ A gS] fp=kkuqlkj ,d nwljs ds lekUrj j[kh gSA X IysV ij Q vkos'k fn;k x;k gSA tcfd nwljh mnklhu jgrh gSA Kkr dhft, % (a) IysV X dh vkarfjd lrg ij i`"B vkos'k ?kuRo (b) IysV ds cka;h rjQ ,d fcUnq ij fo|qr {ks=k , (c) IysV ds e/; ,d fcUnq ij fo|qr {ks=k (d) IysV ds
nka;h rjQ ,d fcUnq ij fo|qr {ks=k
Ans. (a) 24.
Q 2A
Q Q Q (b) 2A towards left (c) 2A towards right (d) 2A towards right 0 0 0
Three identical metal plates with large surface areas are kept parallel to each other as shown in figure. The leftmost plate is given a charge Q, the rightmost a charge – 2Q and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate. cM+s i`"Bh; {ks=kQy dh /kkrq dh rhu leku IysVas fp=kkuqlkj ,d nwljs ds lekUrj j[kh gSA lcls cka;h IysV dks Q vkos'k o lcls nka;h dks – 2Q vkos'k fn;k tkrk gS] tcfd e/; IysV vkos'k jfgr jgrh gSA nka;h IysV dh ckgjh lrg ij
mifLFkfr vkos'k Kkr dhft,A
[Ans. –Q/2]
Ans.
Q 2
manishkumarphysics.in
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