Chapter # 29
Electric Field & Potential
[1]
OBJECTIVE – I 1.
Sol.
2.
Sol.
Fig. shown some of the electronic field lines corresponding to an electric to an electric field. The figure suggests that fuEu iznf'kZr fp=k esa ,d fo|qr {ks=k ls lEc) fo|qr cy js[kk,sa n'kkZ;h xbZ gS -
(A) EA > EB > EC (B) EA = EB = EC (C*) EA = EC > EB (D) EA = EC < EB C Higher separation, Lower electric field. Because Electric field inversly proportional the square of separation. EA = Ec > EB When the seperation between two charges is incresased, the electric potential energy of the charges (A) increases (B) decresaes (C) remains the same (D*) may increase or decrease tc nks vkos'kksa ds e/; nwjh c<+kbZ tkrh gS rks vkos'kksa dh fo|qr fLFkfrt ÅtkZ (A) c<+rh gS (B) ?kVrh gS (C) iwoZor~ jgrh gSA (D*) c<+ ;k ?kV ldrh gS D When the separation between two charges is increased, the electric potential Energy of charge may incease or decrease. If Both charge are like charge then electric potential energy of charge decreases. kq1q2 r If Both charge are unlike charge then electric potential energy of charge increases. U
U
3.
Sol.
4.
kq1q2 r
If a positive charge is shifted from a low-potential region to a high-potential region, the electric potential energy (A*) increases (B) decresaes (C) remains the same (D) may increase or decrease
vxj ,d /kukRed vkos'k dks U;wu foHko {ks=k ls mPp foHko {ks=k dh vksj ys tkrs gS rks fo|qr ÅtkZ (A*) c<+rh gSA (B) ?kVrh gSA (C) ,d leku jgrh gSA (D) ?kV ldrh gS ;k c<+ ldrh gSA
HCV_29_Obj I_3
A Electric Potential Energy = qv = q(vf-vi) If positive charge is shifted from a Low potential region to a High-Potential region, then electric Potential Energy increases. Two equal positive charges are kept at points A and B. The electric potential at the points between A and B (excluding these points) is studied while moving from A to B. The potential (A) continuously increases (B) continuosly decreases (C) increases then decreases (D*) decreases than increases nks leku /kukRed vkos'k fcUnq A o B ij j[ks gSaA A ls B dh vksj pyrs gq, fcUnq A o B ds e/; fo|qr foHko dk v/;;u ¼bu fcUnqvksa ds vfrfjDr½ fd;k tkrk gSA fo|qr foHko (A) yxkrkj c<+rk gS (B) igys c<+rk gS fQj ?kVrk gS (C) yxkrkj ?kVrk gS (D*) igys ?kVrk gS fQj c<+rk gS
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Chapter # 29 Sol.
Electric Field & Potential
A Q
Q
A
r
B
KQ r V Electric Potential KQ KQ KQr VP x r x x r x V
Q A
x
KQ KQ 4KQ pmin r r r 2 2
Sol.
B
r-x
V(Potentail)
Vmin
r = 2x x = r/2
v
Q
P
dvp KQr r 2x 0 2 dx r r x
5.
[2]
r at x 2
x=
r 2
r (distance)
The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting theaxes at points A, B, C and D having coordinates (a,0), (0,a), (-a,0), (0,-a) respectively. Out of the points on the periphery of the circle, the potential is minimum at ewy fcUnq ij fo|qr {ks=k /kukRed X-v{k ds vuqfn'k gS ewy fcUnq dks dsUnz ekurs gq, ,d NksVk o`Ùk [khapk tkrk gS tks v{kksa dks A, B, C o D fcUnqvksa ij dkVrk gSA bu fcUnqvksa ds funs±'kkad Øe'k% (a,0), (0,a), (-a,0), (0,-a) gSA o`Ùk dh ifjf/k ij fLFkr fcUnqvksa esa ls U;wure foHko gksxk y (A*) A (B) B (C) C (D) D A
E E0ˆi Given
B (0,a)
(Potential at A) :-
VA E0ˆi aiˆ E0 9
VB E0ˆi aiˆ 0
x
D (0,-a)
(Potential at C) :-
A (a,0)
C (-a,0)
(Potential at B) :-
VC E0ˆi a ˆi E0a (Potential at D) :-
VD E0ˆi a ˆj 0 6.
Sol.
7.
If a body is charged by rubbing it, is weight (A) remains precisely constant (C) decreases slightly
(B) increases slightly (D*) may increase slightly or may decrease slightly ;fn ,d fudk; dks jxM+dj vkosf'kr fd;k tkrk gS rks bldk nzO;eku (A) fu;r jgrk gSA (B) FkksM+k lk c<+rk gSA (C) FkksM+k lk ?kVrk gSA (D*) FkksM+k lk c<+ vFkok FkksMk+ lk ?kV ldrk gS D If a body is charged by rubbing it, its weight may inc4rease slightly or may decrease slightly. An electric dipole is placed in a uniform electric field. The net electric force on the dipole (A*) is always zero (B) depends on the orientation of the dipole (C) can never be zero (D) dependw on the strength of the dipole ,d fo|qr f}/kzqo dks ,d leku fo|qr {ks=k esa j[kk tkrk gSA f}/kzqo ij dqy fo|qr cy (A*) ges'kk 'kwU; gksrk gS (B) f}/kzqo ds vfHkfoU;kl ij fuHkZj djrk gS (C) dHkh Hkh 'kwU; ugha gks ldrk gSA (D) f}/kzqo ds lkeF;Z ij fuHkZj djrk gS
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Chapter # 29 Sol.
Electric Field & Potential
A
A
E B
-q
E q (i)
8.
Sol.
[3]
B q E E A -q (ii)
Net Electric force = FA + FB = -qE + qE =0 In a uniform Electric field the net Electric force on the dipole is always zero. Consider the situtaion fig. The work done in taking a point charge from P to A is W A, from P to B is W B and from P to C is W C fuEu iznf'kZr fp=k ij fopkj fdft,A fcUnq vkos'k dks P ls A rd ys tkus esa W A P ls B rd ys tkus esa W B rFkk P ls C rd ys tkus esa W C dk;Z djuk iM+rk gS &
(A) W A < W B < W C (C*) W A = W B = W C C
(B) W A > W B > W C (D) none of these
Electric Potential at A ‘due to q’
VA
Kq r
Electric Potetial at B ‘due to q’
VB
Kq r
& Electric potential at c ‘due to q’
VC
Kq r
work done = -u = -qV
C
r
r q
A
r B
Let at 'P ', VP 0
Here VA = VB = VC The work done is taking a point charge from P to A, B & C is same. so wA = wB = wc 9.
Sol.
A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the roatating charge in one complete revolution is (A*) zero (B) positive (C) negative (D) zero if the charge Q is at the centre and nonzero otherwise ,d fcUnq vkos'k q dks ,d vU; fcUnq vkos'k Q }kjk mRiUu fo|qr {ks=k esa ,d o`Ùk ds vuqfn'k ?kqek;k tkrk gSA fo|qr {ks=k esa ,d o`Ùk ds vuqfn'k ?kqek;k tkrk gSA fo|qr {ks=k }kjk ,d iw.kZ pØ esa ?kwf.kZr vkos'k ij fd;k x;k dk;Z gksxk (A*) 'kwU; (B) /kukRed (C) _.kkRed (D) 'kwU; ;fn vkos'k Q dsUnz ij gS vFkok v'kwU;A E A In a circle Electric field due to t ‘Q’ is always perpendicular v v to the displacement oif the charge ‘q’. E Workdone = F.d q r Q qE .d q E.d 0 =
Ed Workdoen by the Electric field on the rotating charge in one complete revolution is zero.
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Chapter # 29
Electric Field & Potential
[4]
OBJECTIVE – II 1.
Sol.
2.
Mark out the correct options. (A*) The total charge of the universe is constant. (B) The total positive charge of the universe is constant. (C) The total negative charge of the universe is constant (D) The total number of charged particles in the universe is constant fuEu esa ls lgh fodYi gS (A*) fo'o dk dqy vkos'k fu;r gSA (B) fo'o dk dqy vkos'k fu;r gSA (C) fo'o dk dqy _.kkRed vkos'k fu;r gSA (D) fo'o esa vkosf'kr d.kksa dh dqy la[;k fu;r gSA A The total charge (Positive + negative) oif the universe is constant. A point charge is brought in an electric field. The electric field at a nearby point
HCV II _Ch-29_Obj.2_2
fo|qr {ks=k esa ,d fcUnq vkos'k dks yk;k tkrk gSA utnhd ds fcUnq ij fo|qr {ks=k
Sol.
3.
Ans.
4.
Sol.
(A) will increase if the charge is positive (B) will decrease if the charge is negative (C*) may increase if the charge is positive (D*) may decrease if the charge is negative (A) c<+sxk vxj vkos'k /kukRed gSA (B) ?kVsxk vxj vkos'k _.kkRed gSA (C*) c<+ ldrk gS vxj vkos'k /kukRed gSA (D*) ?kV ldrk gS vxj vkos'k _.kkRed gSA CD The electric field at a near by point may increase if the charge is positive or may decreases if the charge is negative. The electric field and the electric potential at a point are E and V respectively (A) If E = 0, V must be zero (B) If V = 0, E must be zero (C) If E 0, V cannot be zero (D) If V 0, E cannot be zero fdlh fcUnq ij fo|qr {ks=k ,oa fo|qr foHko Øe'k% E o V gS (A) ;fn E = 0, V Hkh 'kwU; gksxkA (B) ;fn V = 0, E Hkh 'kwU; gksxkA (C) ;fn E 0, V 'kwU; ugha gks ldrkA (D) ;fn V 0, E 'kwU; ugha gks ldrkA None of these buesa ls dksbZ ugha A We not commenton Electric field as well as Electric potential Because Here not given the informatio about the existance of charge in surrounding area. The electric potential decreases uniformly from 120 V to 80 V as one moves on the X-axis from x = – 1 cm to x = + 1 cm. The electric field at the origin (A) must be equal to 20V/cm (B*)may be equal to 20V/cm (C*) may be greater than 20V/cm (D) may be less than 20V/cm x-v{k ds vuqfn'kx = –1 lseh ls x = +1 lseh rd xfr djus ij fo|qr foHko ,d leku :i ls 120 oksYV ls 80 oksYV rd ?kVrk gSA dsUnz fcUnq ij fo|qr {ks=k (A) 20 oksYV@lseh ds cjkcj gksuk pkfg, (B*) 20 oksYV@lseh ds cjkcj gks ldrk gSA (C*) 20 oksYV@lseh ls T;knk gks ldrk gSA (D) 20 oksYV@lseh ls de gks ldrk gSA BC E E
V E dr (v f vi ) E dr Ex B A
80 120 Ex 2 Ex
x=-1 A V=120 v
E x=1 B V=80 v
40 20 v / cm 2
If electric field lines lies in ‘x’ direction than it may be equal to 20 v/cm. If Electric field lines liesin ‘x-y’ direction than it may be greater than 20 v/cm. 5.
Which of the following quantites do not depend on the choice of zero potential or zero potential energy (A) potential at a point (B*) potential difference between two points (C) potential energy of a two - charge system (D*) change in potential energy of a two-charge system
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Chapter # 29
Sol.
Sol.
[5]
fuEu esa ls dkSulh jkf'k;k¡ 'kwU; foHko ;k 'kwU; fLFkfrt ÅtkZ ds p;u ij fuHkZj ugha djrh gS (A) ,d fcUnq ij foHko (B*) nks fcUnqvksa ds e/; foHkokUrj (C) nks vkos'kksa ds fudk; dh fLFkfrt ÅtkZ (D*) nks vkos'kksa ds fudk; dh fLFkfrt ÅtkZ esa ifjorZu CD
Vp 6.
Electric Field & Potential
Q
KQ KQ 0 r r
P r
-Q r
An electric dipole is placed in an electric field generated by a point charge (A) The net electric force on the dipole must be zero (B) The net electric force on the dipole may be zero (C) The torque on the dipole due to the field must be zero (D*) The torque on the dipole due to the field may be zero ,d fo|qr f}/kzqo ij fcUnq vkos'k }kjk mRiUu fo|qr {ks=k esa j[kk gS (A) f}/kzqo ij dqy oS|qr cy 'kwU; gksuk pkfg,A (B) f}/kzqo ij dqy oS|qr cy 'kwU; gks ldrk gSA (C) {ks=k ds dkj.k f}/kzqo ij cy vk?kw.kZ 'kwU; gksuk pkfg,A (D*) {ks=k ds dkj.k f}/kzqo ij cy vk?kw.kZ 'kwU; gks ldrk gSA D In the uniform Electric field the net electric force on the dipole is alwas zero. In uniform Electric field the torque on the dipole due to field may be zero. q
E
E
-q (i) T=0 Here Torque T = 2 qEl sin 0
E
q -q
(ii)
7.
A proton and an electron are placed in a unifrorm electric field. (A) The electric forces acting on them will be equal (B*) The magnitudes of the forces will be equal (C) Their acelerations will be equal (D) The magnitudes of the accelerations will be equal. ,d izksVkWu o ,d bysDVªkWu ,d le:i fo|qr {ks=k esa j[ks x, gSa (A) mu ij yxus okyk oS|qr cy cjkcj gksaxsA (B*) cyksa ds ifjek.k cjkcj gksaxsA (C) muds Roj.k cjkcj gksaxsA (D) muds Roj.k ds ifjek.k cjkcj gksaxsA Sol. B Proton contain positive charge = 1.6 × 10-19 = e electron contain negative charge = -1.6 ×10-19 = -e force inoniforma electric field is = qE The Magniludes of Electric force is equal But direction of electric force is opposite to each other. mass of proton = 1.67 ×10-27 Kg. mass of electron = 9.1 × 10-31 Kg. So that Magnitudes of their acceleration will be unequal. 8.
Sol.
The electric field in a region is directed outward and is propertional to the distance r form the origin. Taking the electric potential at the origin to be zero (A) it is uniform in the region (B) it is proportional to r 2 (C*) it is proportional to r (D) it increases as one goes away from the origin ,d {ks=k esa fo|qr {ks=k ckgj dh vksj gS rFkk ewy fcUnq ls nwjh r lekuqikrh gSA ewy fcUnq ij fo|qr foHko dk eku 'kwU; ysus ij (A) foHko {ks=k esa ,d leku gSA (B) foHko r ds lekuqikrh gSA (C*) foHko r2 ds lekuqikrh gSA (D) foHko ewy fcUnq ls nwj tkus ij c<+rk gSA C V = -E.dr Given (E r) (V - 0) = -E.dr E r2
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