16-Electrostatics
TOPIC
ELECTROSTATICS
16
SECTION - I : STRAIGHT OBJECTIVE TYPE
[k.M-I: lh/ks oLrqfu"B izdkj
16.1 A point charge + Q is placed at the centroid of an equilateral triangle. When a second charge + Q is placed at a vertex of the triangle, the magnitude of the electrostatic force on the central charge is 8 N. The magnitude of the net force on the central charge when a third charge + Q is placed at another vertex of the triangle is:
,d fcUnq vkos'k + Q leckgq f=kHkqt ds dsUnzd ij j[kk gqvk gSA tc ,d nwljk vkos'k + Q f=kHkqt ds 'kh"kZ ij j[kk tkrk gS rks oS|qrh; cy tks dsUnzh; vkos'k ij yx jgk gS dk ifjek.k 8 N gSA ;fn rhljk + Q vkos'k f=kHkqt ds vU; 'kh"kZ ij j[k fn;k tk;s rks dsUnzh; vkos'k ij yxus okyk cy gksxkA
(A) zero (B) 4 N (C) 4 N (D*) 8 N
Sol. R = = 8 N
16.2 The electric field inside a sphere which carries a volume charge density proportional to the distance from the origin = r ( is a constant) is : mlxksysdsvUnjfo|qr{ks=kD;kgksxkftldkvk;ruvkos'k?kuRoewyfcUnqlsnwjhdslkFk=rlsfn;ktkrk gSA(tgk¡,dfu;rkadgSA)
r3
(A) 4 (B*)
r2
4 (C)
r2
3 (D) noneof these buesa ls dksbZ ugha
Sol. We can consider all the charge inside the sphere to be concentrated on the center of sphere Consider an elementry shell of radius x and thickness dx. gedYiukdjrsgSfdxksysdsvUnjmifLFkrvkos'k]xksysdsdsUnzijbdV~BkgSA
x f=kT;krFkkdxeksVkbZdsdks'kdhdYiukdjrsgSA
E = Kdq
r 2
r
K4x2dx (x)
= 0 =
r 2
k4 r 2
r
x3dx =
0
r 2
40
16.3 A particle of charge - q & mass m moves in a circle of radius r around an infinitely long line charge of linear charge density + . Then time period of revolution of charge will be :
M.Bank_E.S._3.7
,d–qvkos'komnzO;ekudk,dd.k+ jsf[kdvkos'k?kuRookysvuUryEckbZds,djsf[kdvkos'kdspkjksa vksjrf=kT;kdso`ÙkesaxfrdjjgkgSAvkos'kds?kw.kZudkvkorZdkygksxk&
(A*) T = 2 r
1
2 k q
(B) T2 =
4 2 m
r
2 k q
1
(C) T =
2 r
m (D) T =
1
2 r
where tgkW k = 4 0
Sol. From the F.B.D. of charge, we have :
vkos'kdseqDroLrqfp=kls
2k
mv2
q =
r
so vr% v =
2r
Now vc, T = v
= 2r
where, ;gk¡ k =
1
40
16.4 An infinitely long plate has surface charge density . As shown in the fig, a point charge q is moved from A to B. Net work done by electric field is:
,dvuUryEckbZdhIysVdki`"Bh;vkos'k?kuRogSAfp=kesafn[k ;svuqlkj,dfcUnqorvkos'kqdksAls Brdys
tkrsgSaAfo|qr{ks=k}kjkfd;kx;kdqydk;ZgS&
q q q q
(A*) 2 0 (x1 x2) (B) 2 0 (x2 x1) (C) 0 (x2 x1) (D) 0 (2 r + r)
Sol. W =
→ where ( ;gk¡ → = ˆi & → (x
– x )ˆi ) W
= q .(x
– x )
net
qE. d
E 2 0
d 1 2
net
2 0 1 2
16.5 Figure shows two large cylindrical shells having uniform linear charge densities + and – . Radius of inner cylinder is ‘a’ and that of outer cylinder is ‘b’. A charged particle of mass m, charge q revolves in a circle of radius r (where a < r < b). Then it's speed ‘v’ is : (Neglect gravity and assume the radii of both the cylinders to be very small in comparison to their length.) fp=kesankscM+scsyukdkjdks'kftudk,dlekujs[kh;vkos'k?kuRo+o–gS]fn[kk;sx;sgSAvkUrfjdcsyu rFkkckácsyudhf=kT;karFkkbgSAnzO;eku m vkSj vkos'kqdk,dvkosf'krd.krf=kT;k (tgk¡a< r< b). ds o`Ùkesa?kwerkgSArksbldhpkyvgksxhA(xq:Rodksux.;ekfu;s]nksuksacsyuksadhf=kT;k,amudhyEckbZdhrqyuk esacgqrNksVhgSA)
(A*)
(B)
2k
(C)
(D) (D)
Sol. Electric field between the two cylinders = r
Force on charge q = This force is centripetal force
2kq r
2kq
r
mv2
= r
v = =
gy nksaukscsyuksadschpfo-{ks=k= 2k
q vkos'k ij cy = 2kq
;gcyvfHkdsUnzh;cygSA
2kq r
mv2
= r
v = =
16.6 Figure shows three circular arcs, each of radius R and total charge as indicated. The net elecric potential at the centre of curvature is :
rhu o`Rrkdkj pki] izR;sd dh f=kT;k R ,oa mu ij dqy vkos'k fp=k esa n'kkZ;s x;s gSA oØrk dsUnz ij ifj.kkeh foHko gS
+Q
–2Q
45°
•
R
30°
+3Q
Q
(A*) 20R
Sol. V = V1 + V2 + V3
(B)
Q
40R
2Q
(C) 0R
Q
(D) 0R
(where V1, V2 & V3 are potentials due to the three parts of ring)
tgk¡V1,V2oV3oy;dsrhuksaHk xksadsdkj.kfoHkogS
1 Q 1
2Q
1 3Q
1 2Q Q
= 4 .
+ 4
R +
4
R =
4 . R =
2 R
0 R 0
0
0 0
16.7 An electric field is given by Ex (2, 4) is taken as zero, is
= –2x3 kN/C. The potential of the point (1, –2), if potential of the point
fo|qr {ks=k E = –2x3 kN/C }kjk O;ä fd;k tkrk gSA ;fn fcUnq (2, 4) dk foHko 'kwU; ekusa] rks fcUnq (1, –2) dk foHko gksxk &
(A) – 7.5 × 103 V (B*) 7.5 × 103 V (C) – 15 × 103 V (D) 15 × 103 V
→
Sol. dV = –
→ = – (2x3 103 ˆi ).(dx ˆi dy ˆj dz kˆ)
= 2x3 ×103 dx
E.dr
v 1
dV = – ( 2x3) × 103 dx V = 7.5 × 103 V
0 2
16.8 Two concentric uniformly charged spheres of radius 10 cm & 20 cm are arranged as shown in the figure. Potential difference between the spheres is: nksladsUnzh;,dlekuvkosf'krxksyksadhf=kT;k,sa10cmo20cmgS]budksfp=k uqlkjO;ofLFkrfd;kx;kgSAxksyksa dschpfoHkokUrjgSA
(A*) 4.5 1011 V (B) 2.7 1011 V (C) 0 (D) noneof thesebuesalsdksbZugha
Sol. Potential difference due to inner 10C charge
vkUrfjd10Cvkos'kdsdkj.kfoHkokUrj
1 1
= K 10 .1 .2 = 9 × 1010 (5) = 45 × 1010 = 4.5 × 1011V
K 20 – K 20
Potential difference due to outer charge =
0.2
0.2
= 0 V
ckávkos'kdsdkj.kfoHkokUrj=0
P.d. = 4.5 × 1011V
16.9 Figure shows an electric line of force which curves along a circular arc. The magnitude of electric field intensity is same at all points on this curve and is equal to E. If the potential at A is V, then the potential at B is :
fp=kesafo|qrcyjs[k fn[k bZzxbZgStkso`Rrh;pkidsvuqfn'kgSAfo|qr{ks=kdhrhozrkoØdsizR;sdfcUnqij lekugSvkSjEdscjkcjgSA;fnAijfoHkoVgSrksBijfoHkogksxkA
(A*) V – ER (B) V – 2ER sin 2
(C) V + ER (D) V + 2ER sin 2
Sol. VA
– VB
= work done by electric field on + 1 coul. charge from A to B = E R
++1 dwykWevkos'kijAlsBrdfo-{ks=k}kjkfd;kx;kdk;ZVA –VB = ER
V = V – ER = V – ER
16.10 Figure shows a solid hemisphere with a charge of 5 nC distributed uniformly throughout its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the centre of curvature at distance 15 cm. The electric potential at point P due to the hemisphere, is :
fp=kesa,dv)ZBkslxksykn'kkZ;kx;kgSftlij5nCvkos'kvk;ruesa,dleku:ilsforfjrgSAv)Zxksyk] leryijj[k gSrFk leryesaf=kT;h;js[k dsvuqfn'koØrkdsUnzls15lseh-dhnwjhijfcUnqPfLFkrgSAfcUnq P ijv)Zxksysdsdkj.kfo|qrfoHkogS&
15cm
(A) 150 V (B*) 300 V (C) 450 V (D) 600 V
Sol. By argument of symmetry, it will be half of the potential produced by the full sphere
lfeferhdsrdZij;giwjsxksys}kjkmRiUufoHkodkvk/k gksxk
Charge on hemisphere = Q, so charge on sphere = 2Q
v)Zxksysijvkos'k=Q,blfy,xksysijvkos'k=2Q
1 . K (2Q) KQ
2 R R
KQ 9 109 5 109
V = R =
15 102
= 300 V
16.11 A point charge Q is placed at a distance d from the centre of an uncharged conducting sphere of radius
R. The potential of the sphere is (d > R) :
,dvukosf'krRf=kT;kdspkydxksysdsdsUnzlsd(d> R)nwjhij,dfcUnqor~vkos'kQj[k tkrkgSAxksysdk foHkogS&
(A)
1
4 0
. Q
(d R)
(B*)
1 . Q
4 0 d
(C)
1 . Q
4 0 R
(D) zero 'kwU;
Sol. Net potential of the sphere due to the induced charge is zero. Therefore potential is due to the point charge only, equal to potential at the centre of the sphere.
gy çsfjrvkos'kdsdkj.kxksysdkifj.k ehfoHko'kwU;gSAvr%foHkodsoyfcUnqorvkos'kdsdkj.kgksxk]tksfddsUnz ijfoHkogksxkA
Q
+
16.12 Two point dipoles of dipole moments pkˆ
& p kˆ are located at (0, 0, 0) & (1 m, 0, 2 m) respectively.
2
The resultant electric field due to the two dipoles at the point (1 m, 0, 0) is : nksfcUnqor~f}&/kzqoftudsf}/kzqovk?kw.kZ pkˆ vkSj p kˆ gS;sØe'k%(0,0,0)rFk (1m,0,2m)ijfLFkrgSA fcUnq(1m,0,0)ijnksuksaf}/kzqoksadsdkj.kifj.kehfo|qr{ks=kgSA
9 p
(A) 32
kˆ (B*) 7 p 32 0
kˆ (C) 7 p 32 0
kˆ (D) noneof thesebuesalsdksbZugha
Sol.
→
The given point is at axis of P
dipole and at equatorial line of
→ dipole so total field at given point is.
P
→
P →
fn;kx;kfcUnqf}/kzqo 2 dhv{krFk f}/kzqo P dhfuj{kjs[k ijfLFkrgSAvr%fn;sx;sfcUnqijdqy{ks=k
→
→ kP
→
2k(P / 2)
→ 1 1
→
7P
E = –
(1)3 +
(2)3
= k
= 32
16.13 A dipole of dipole moment p is kept at the centre of a ring of radius R and charge Q. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole is:
f}/kzqo vk?kw.kZ p dk f}/kzqo Q vkos'k ;qDr R f=kT;k dhs oy; ds dsUnz ij j[kk gSA f}/kzqo vk?kw.kZ dh fn'kk oy; dh v{k ds vuqfn'k gSA f}/ kzqo ds dkj.k oy; ij ifj.kkeh cy gS :
(A) zero 'kwU; (B*)
kPQ
kPQ R3
(C)
2kPQ R3
(D)
R3
kPQ R3
only if the charge is uniformly distributed on the ring.
dsoy rHkhtc ;fn vkos'k oy; ij le:i forfjr gksA
k p
Sol. Electric field at each point on the surface of ring due to dipole is E =
R
in direction opposite to the dipole moment. (figure below)
Hence net force on ring is F = QE =
Alternate solution
kpQ R3
Electric field due to ring at point P on its axis at a distance x from centre O of ring is
E = k .
Qx
(x2 R2 )3 / 2
dE
; dx at x0
kQ
= R3
Force on dipole =
dE dx =
k Qp R3
k p
gy f}/kzqodsdkj.koy;dhlrgijfLFkrçR;sdfcUnqijfo|qr{ks=kE= R3
f}/kzqovk?kw.kZdsfoifjrfn'k esa(uhpsfp=k)
vr% oy; ij dqy cy F = QE =
vU;fof/k
kpQ R3
oy;dsdkj.kbldhv{kijdsUnzOlsx nwjhijfLFkrfcUnqPijfo-{ks=k
E = k .
Qx
(x2 R2 )3 / 2
dE
; dx at x0
kQ
= R3
f}/kzqo ij cy =
dE dx =
k Qp R3
16.14 A charge Q is placed at a distance of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is . Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface taking direction of area vector along outward normal as positive, is
,dvkos'kQ,Rf=kT;kdhpdrhdsdsUnzls4Rnwjhijj[kkx;kgSApdrhlsxqtjusokykdqy¶yDl gSAvcR f=kT;kdkv)Zxksykdkjdks'k]pdrhijblçdkjj[k tkrkgSfd;gcUni`"BcukrkgSAoØi`"Blsxqtjusokys¶yDl dkekugksxkA({ks=kQylfn'kdksi`"BdsckgjdhrjQ/kukRedekfu,)
(A) zero (B) (C*) – (D) 2
Sol. After covering with a hemispherical shell ;
disc
= 0 (from Gauss law)
v)Zxksyh; dks'k dks ifjc) djus ds i'pkr~; + =0(xkÅl ds fu;e ls)
shell
disc
= –
dks'k
pdrh
16.15 A conducting disc of radius R rotates about its axis with an angular velocity .Then the potential difference between the centre of the disc and its edge is (no magnetic field is present) :
R f=kT;kdh,dpkydpdrhbldsv{kds lkis{kdks.kh;osxls?kwerhgSrkspdrhdsdsUnzrFkkfdukjsds e/; foHkokUrj gksxk (dksbZ pqEcdh; {ks=k mifLFkr ugha gS) :
(A) zero (B*)
me2 R2
2 e
(C)
meR3 3e
em R2
(D) e
2
Sol. eE = me r (Balancing of forces on e–)
e– ij cy larqfyr djus ij
0 R m 2 R
Also, VO
– VR = –
E.dr E.dr =
e r dr e
R 0 0
So, blfy, V =
me2R2 2e
16.16 At distance ' r ' from a point charge, the ratio represented by :
u
v 2 (where ' u ' is energy density and ' v ' is potential) is best
u
,dfcUnqvkos'kls'r'nwjhij v 2 dknwjhdslkFkifjorZufuEuesalsfdlxzkQ}kjkiznf'kZrgksxkA(tgk¡uÅtkZ
?kuRo rFkk 'v' foHkokUrj gS)
(A)
(B*)
(C)
(D)
Sol. u = 1 2
E2 = 1
0 2
0K2Q2
2r 4
v = KQ
r
u
=
2 0K
2 Q2
r 4
= 1 0
v 2 K 2Q2
r 2
2 r 2
u 1
becauseD;ksafd
v 2 r 2
so the correct option is B.
blfy, lgh dFku B gSA
16.17 A ring of radius R is placed in the plane with its centre at origin and its axis along the x-axis and having uniformly distributed positive charge. A ring of radius r (<< R) and coaxial with the larger ring is moving along the axis with constant velocity then the variation of electrical flux () passing through the smaller ring with Position will be best represented by:
,dRf=kT;kdh,doy;dks,dryesaj[kkx;kgSAbldkdsUnzewyfcUnqijrFk v{kx-v{kdsvuqfn'kgSAblij
,dleku/kukRedvkos'kforfjrgSA,dr(< r1 Therefore E1 > E2
vRk%E1 > E2
i.e. Net field at A is towards a2.
vFk ZrAijifj.k ehfo|qr{ks=ka2dhrjQgksxkA
K.(r1)
V1 =
V2 =
r1
K.(r2 )
r2
= K
= K V
= V2 .
16.38 Two infinite, parallel, non–conducting sheets carry equal positive charge density . One is placed in the yz plane at x = 0 and the other at x = a. Take potential V = 0 at x = 0.
nksvuUr]lekUrj] vpkydifêdkvksaijcjkcj/kukRedvkos'k?kuRogSA,ddksy-z ryesarFkknwljhdksx
= a ij j[krs gSaA x = 0 ij foHko V = 0 ekusaA
(A*) For 0 x a, potential V
= 0. (B*) For x a, potential V
(x a)
0
(C) For x a, potential V =
(x a)
0
(D*) For x 0 potential V = x
0
(A*) 0 x a ds fy,] foHko V = 0. (B*) x a ds fy,] foHko V = (x a)
x x 0
(C) x a ds fy,] foHko V = (x a)
(D*) x 0 ds fy,] foHko V = x
x 0 x 0
x
Sol. 0 x a : V
= Exdx V(0)
= 0 (as E = 0)
x 0
x
x
x
x a ; V
= Exdx V(a)
=
dx V(a) =
(x – a) ; (As, D;ksafdV
x
a
= 0)
x
a 0 0 a
x 0 ; V =
Exdx V(0)
=
.x
+ V(0) =
.x
; (As, D;ksafdV0 = 0)
0 0 0
16.39 Two concentric rings of radii R1 = m and R2 = 4m are placed in y-z plane with their centres at origin.
They have uniform charge –q and +Q = 2 2 q on the inner and outer rings respectively. Consider the electrostatic potential to be zero at infinity. Then
f=kT;kvksaR1= 6 m vkSjR2=4m dhladsUnzh;oy;sy-zryesafLFkrgS]mudsdsUnzewyfcUnqijgSaAvkUrfjd vkSjckáoy;ksa ij,d leku vkos'kØe'k%–qvkSj+Q=2 2 qgSAvuUrijfo|qrfoHko'kwU;gSArks
(A) The electric potential is zero at origin.
(B*) The electric field intensity is zero at r = 2 m.
(C*) A positive charged particle disturbed from origin along the x-axis will restore back to origin. (D*) Where potential is maximum on the x-axis, field intensity is zero. (A)ewyfcUnqijfo|qrfoHko'kwU;gSA
(B*)r=2mijfo|qr{ks=kdhrhozrk'kwU;gSA
(C*)x-v{kdsvuqfn'kewyfcUnqlsFk sM+kfoLFk firdjusij,d/kukosf'krd.kewyfcUnqijokilvktk;sxkA (D*) x-v{kijtgk¡foHkovf/kdregS]fo|qr{ks=kdhrhozrk'kwU;gSA
Sol. (Moderate) V at origin 0
ewyfcUnq ij V 0
E(r = 2 m) =
K(q)r K.Q.r
(R2 r 2 )3 / 2 (R2 r 2 )3 / 2
= K.rq
1
103 / 2
23 / 2.103 / 2 = 0
From origin to r = 2, field is towards origin.
ewyfcUnqlsr=2,{ks=kewyfcUnqdhvksjgSA
16.40 In front of an earthed conductor a point charge + q is placed as shown in figure :
i`FohlstqM+s,dlqpkyddslEeq[k,dfcUnqvkos'k+qfp=kkuqlkjj[k gS&
(A*) On the surface of conductor the net charge is negative.
pkyd dh lrg ij dqy vkos'k _.kkRed gSA
(B*) On the surface of conductor at some points charges are negative and at some points charges may be positive distributed non uniformly
pkyddhlrgijdqNfcUnqvksaij_.k RedrFk dqNfcUnqvksaij/kukRedvkos'kvleku:ilsforfjrgksldrsgSaA
(C*) Inside the conductor electric field due to point charge is non zero
pkyddsvUnj]blfn,x,fcUnqvkos'kdsdkj.kfo|qr{ks=k'kwU;ughagSA
(D) None of these
mijksäesalsdksbZughaA
Sol. Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside. Field due to only q is non-zero. pkydijvkos'kblizdkjforfjrgksxkfdblvkos'k,oackgjhvkos'kdsdkj.kpkyddsvUnjfo|qr{ks=k'kwU; gksxkAdsoyqdsdkj.kfo|qr{ks=kv'kwU;gSA
SECTION - III : ASSERTION AND REASON TYPE
[k.M- III: dFku vkSj dkj.k izdkj
16.41 Statement 1 : Electric field E at a point P is zero if potential at that point is zero.
Statement 2 : Potential difference between two points in space is zero if electric field at all points in space is zero.
oDrO;1:,dfcUnqPijfo|qr{ks=kE'kwU;gS;fnmlfcUnqijfoHko'kwU;gSA oDrO;2:fdlh{ks=kesanksfcUnqvksadse/;foHkokUrj'kwU;gS;fn{ks=kesalHkhfcUnqvksaijfo|qr{ks=k'kwU;gSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1lR;gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dklgh Li"Vhdj.kugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1lR; gS, oDrO;-2vlR; gSA
(D*) Statement-1 is False, Statement-2 is True.
oDrO; -1vlR; gS, oDrO;-2lR; gSA
Sol. Potential at a point is zero does not imply that electric field at same point should be zero. For instance in the equatorial plane of a dipole, potential at any point is zero but electric field is not zero. Hence statement 1 is false.
No electric field in space means , potential at all points in space is same. Hence potential difference between any two points is zero. Hence statement 2 is true.
Statement–1 is False, Statement–2 is True. foHkofdlhfcUnqij'kwU;gS;giznf'kZrughadjrkgSfdfo|qr{ks=kHkhmlhfcUnqij'kwU;gksxkAf}/kzqodsfuj{kh;ryesa] foHkofdlhHkhfcUnqij'kwU;gksrkgS]ysfdufo|qr{ks=k'kwU;ughagksrkgSAvr%oDrO;1vlR;gSAfdlhLFkuesafo|qr
{ks=kdkugksuk];giznf'kZrdjrkgSfdmlLFkuesalHkhfcUnqvksaijfoHkolekugSAblfy;sfdUghanksfcUnqvksadschp foHkokUrj'kwU;gksxkAvr%oDrO;2lR;gSA
dFku -1vlR; o dFku -2lR; gSA
16.42 STATEMENT-1 : A point charge q is placed at centre of spherical cavity inside a spherical conductor as shown. Another point charge Q is placed outside the conductor as shown. Now as the point charge Q is pushed away from conductor, the potential difference (VA – VB) between two points A and B within the cavity
of sphere remains constant.
oDrO;-1:fn[k ;sx;sxksyh;pkyddsvUnjxksyh;xqfgdk(cavity) dsdsUnzij,dfcUnqvkos'kqfLFkrgSAnwljkfcUnq vkos'kQfn[k ;svuqlkjpkyddsckgjfLFkrgSAvcfcUnqvkos'kQdkspkydlsckgjnwjdhvksj/kdsyusijHkhxksys dhxqfgdkdsvUnjnksuksafcUnqvksaArFk BdschpfoHkokUrj(VA–VB)fu;rjgrkgSA
STATEMENT-2 : The electric field due to charges on outer surface of conductor and outside the conductor is
zero at all points inside the conductor.
oDrO;-2:pkyddhckgjhlrgijmifLFkrvkos'k sarFk pkyddsckgjmifLFkrvkos'k sadsdkj.kla;qDr:ilspkyd dsvUnjlHkhfcUnqvksaijfo|qr{ks=k'kwU;gksxkA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
(A*) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
(C) oDrO;-1lR; gS] oDrO;-2vlR; gS ;
(D) oDrO;-1vlR; gS] oDrO;-2lR; gS
Sol. The electric field inside the cavity depends only on point charge q. Hence VA – VB remains constant even if point charge Q is shifted. Here statement 2 is correct explanation of statement 1.
xqfgdk(cavity)dsvUnjfo|qr{ks=kdsoyfcUnqvkos'kqijfuHkZjdjrkgSA;fnfcUnqvkos'kQdksfoLFk firHkhfd;k tk;] rks ;gk¡ VA–VB fu;r jgrk gSA ;gk¡ oDrO;-2,oDrO;-1dk lgh Li"Vhdj.k gSA
16.43 STATEMENT-1 : For a non-uniformly charged thin circular ring with net charge zero, the electric field at any point on axis of the ring is zero.
STATEMENT-2 : For a non-uniformly charged thin circular ring with net charge zero, the electric potential at each point on axis of the ring is zero.
oDrO;-1 : vleku:i ls vkosf'kr ,diryh oy;ftl ijusV vkos'k'kwU; gS]ds fy, oy; dh v{k ij fdlh fcUnq ijfo|qr{ks=k'kwU;gksxkA
oDrO;-2: vleku:ilsvkosf'kr ,diryh oy;ftlijusVvkos'k'kwU; gS]ds fy,oy; dhv{k ijizR;sd fcUnq ijfo|qrfoHko'kwU;gksxkA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D*) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. For a non-uniformly charged thin circular ring with net zero charge, electric potential at each point on its axis is zero. Hence electric field at each point on its axis must be perpendicular to the axis. Therefore statement 1 is false and statement 2 is true.
,dvleku:ilsvkosf'kriryho`Ùkkdkjoy;ftlijusVvkos'k'kwU;gS]dsfy,bldsv{kijizR;sdfcUnqij foHko'kwU;gksrkgSAblfy;sbldhv{kijizR;sdfcUnqijfo|qr{ks=kv{kdsyEcor~gksxkAblfy;soDrO;-1vlR;
gSrFkkoDrO;-2lR;gSA
16.44 STATEMENT-1 : A uniformly charged disc has a pin hole at its centre. The electric field at the centre of the disc is zero.
STATEMENT-2 : Disc can be supposed to be made up of many rings. Also electric field at the centre of uniformly charged ring is zero..
oDrO;-1: ,dleku :i ls vkosf'kr pdrh ds dsUnz ij fiu (pin) ds vkdkj dk fNnz gSA pdrh ds dsUnzij fo|qr
{ks=k'kwU;gSA
oDrO;-2: pdrhdksdbZlkjhoy;ksalscuhgqbZekuldrsgSrFk ,dleku:ilsvkosf'kroy;dsdsUnzijfo|qr
{ks=k'kwU;gksrkgSA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO; -1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. The electric field due to disc is superposition of electric field due to its constituent ring as given in statement-2. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-2dsvuqlkjfMLddsdkj.kfo|qr{ks=k]?kVdvo;oksoy;ksadsdkj.kfo|qr{ks=kdsv/;kjksi.klsekuldrs gSA oDrO;-1 lR; gS oDrO;-2 lR; gSA oDrO;-2, oDrO;-1 dh lgh O;k[;k djrk gSA
16.45 STATEMENT-1 : Total work done by non uniform electric field on a charged particle starting from rest till any time is non negative. (assume no other forces act on the charged particle)
STATEMENT-2 : The angle between electrostatic force and velocity of the charged particle released from rest in non uniform electric field is always acute. (assume no other forces act on the charged particle)
oDrO;-1: fojkelsizkjEHkdjrsgq;sfdlhHkhle;rd,dvkosf'krd.kijvlekufo|qr{ks=k}kjkfd;kx;kdqy dk;Zv_.k RedgksrkgSD;ksafd¼;gekfu;sfdvkosf'krd.kijdksbZvU;cyughgSA½
oDrO;-2: vlekufo|qr{ks=kesafojkelsNksM+sx;svkosf'krd.kdsosxrFk fLFkjfo|qrdh;cydschpdkdks.k ges'k U;wudks.kgksrkgSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C*) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. From work energy theorem
Final K.E. = Initial K.E. + work done by non-uniform electric field.
initial K.E. = 0 and final K.E. cannot be negative.
Work done by non-uniform electric field on a charged particle starting from rest is non-negative.
Hence statement is true.
Consider a situation in which two point charges +Q are fixed some distance apart. At some distance left of equilibrium point O, a charge +q is released from rest. After the charge +q moving towards right crosses O, it experiences a force towards left.
Hence statement is false.
+Q +q O
fixed charge
Statement-1 is True, Statement-2 is False.
gy% dk;Z&ÅtkZizes;ls]
+Q
fixed charge
vfUre xfrtÅtkZ = izkjfEHkdxfrt ÅtkZ+vlekufo|qr {ks=k}kjk fd;kx;k dk;Z
izkjfEHkdxfrt ÅtkZ= 0rFkk vfUre xfrt ÅtkZ _.kkRed ugh gksldrh gSA
fojkelsizkjEHkdjrsgq;s,dvkosf'krd.kijvlekufo|qr{ks=k}kjkfd;kx;kdk;Zv_.k RedgksxkA blfy;s oDrO; lR; gSA
,d fLFkfr dks yhft, ftlesa nks fcUnq vkos'k +Q dqN nwjh ij fLFkj
gSaA lkE;oLFkk fcUnq O ds cka;h vksj dqN nwjh ij] ,d vkos'k +q dks fojke ls NksM+k tkrk gSA nka;h vksj xfr djrs gq;s vkos'k +q ds O ds ikj djus ds ckn] ;g cka;h vksj ,d cy vuqHko djrk gSA vr% oDrO; vlR; gSA
oDrO;-1 lR; gS] oDrO; -2 vlR; gSA
SECTION - IV : COMPREHENSION TYPE
[k.M- IV:cks/ku izdkj
Paragraph for Question Nos. 19 to 20
iz'u 19 ls20 dsfy, vuqPNsn
A metal ball (Neutral) with radius r is concentric with hollow metal sphere of radius ‘R’, having charge ‘Q’ as shown in figure, Now ball is connected with a very long wire to earth. Then :
rf=kT;kdh,d/kfRod¼mnklhu½xsan‘R’f=kT;kds[k s[kys/kfRodxksysdslkFkladsfUnz;gS,rFk blxksysijfp=k uqlkj ‘Q’vkos'kgSAvcxsandkscgqryEcsrkj}kjki`FohlstksM+ktkrkgS]rc:
16.46 Potential difference between sphere and metal ball, after grounding is :
HkwlaEifdZrdjusdsi'pkr~xksysrFk /kfRodxsandse/;foHkokUrjgksxk%
kQ kQr
kQ r
kQ R
Ans. (c)
(A)
R (B)
R2
(C*)
R 1 R
(D)
r 1 r
16.47 After grounding : HkwlEifdZrdjusdsi'pkr~&
(A) net electric field between sphere and ball is zero.
xksysrFk xsandse/;ifj.kehfo|qr{ks=k'kwU;gksrkgSA
(B) electric field between ball and sphere is zero due to ball only.
xksysrFk xsandse/;fo|qr{ks=kdsoyxsandsdkj.kgksrkgSA
(C*) electric field between sphere and ball due to ball is non–zero.
Ans. (c) Sol.
xksys rFkk xsan ds e/; xsan ds dkj.k fo|qr {ks=k v'kwU; gksrk gSA
(D) electric field between sphere and ball is non-zero, due to sphere
xksys rFkk xsan ds e/; xksys ds dkj.k fo|qr {ks=k v'kwU; gksrk gSA
Vball 0
KX KQ 0
r R
x Qr
R
Potential difference foHkokUrj V V
KQ Kx
s b R R
because potential difference depend only on charge on inner surface after electrostatic condition is reached after grounding.
D;ksafdHkwlEifdZrdsi'pkr~fLFkjoS|qfrdhdhvoLFk esafoHkokUrjdsoyvkUrfjdlrgijmifLFkrvkos'kijfuHkZjdjrk Sg
V V
KQ
r
s b R 1 R
Comprehension – 2
Two positive point charges A and B have charge +q and +2q; mass m and 2m respectively as shown. Both the charges are released from rest when they are at a distance 𝑙0 apart. Neglect gravity and also assume the only force acting on either charge is the electrostatic force due to each other.
nks /kukRed fcUnq vkos'k A rFkk B ij Øe'k% +q rFkk +2q vkos'k gS rFkk budk nzO;eku m rFkk 2m gS tSlk fd fp=kkuqlkj fn[kk;k x;k gSA nksuksa vkos'kksa dks fojke ls NksM+k tkrk gS tc muds chp dh nwjh 𝑙0 gksrh gSA xq:Ro dks ux.; ekfu;s rFkk fdlh Hkh vkos'k ij nwljs ds dkj.k yxus okyk cy dsoy fLFkj fo|qrdh; cy ekfu;sA
16.48 The speed of charge A at the instant separation between both charges is 2𝑙0 is :
tcnksuksavkos'k sadschpdhnwjh2𝑙0 gS]rksml{k.kijvkos'kAdhpky&
(A)
(B)
(C)
(D*)
Sol. (Moderate) Let the speed of charges A and B be VA and VB when the separation between them is 𝑙o.
Then from conservation of momentum
–mVA +2 mVB = 0 or VA = 2 VB
Applying conservation of energy, as the separation increases from 𝑙o to 2𝑙o.
Gain in K.E of system of charges = Loss in electrostatic potential energy of the system of charges.
ekukvkos'kArFkkBdhpkyVA rFkkVB gS]tcmudschpdhnwjh𝑙o gSArkslaosxlaj{k.kdsfu;els]
–mVA +2 mVB = 0 ;k VA = 2 VB
tcnwjh𝑙o ls2𝑙o c<+rhgSrksÅtkZ&laj{k.kdkfu;eyxkusij] vkos'k sadsfudk;dhxfrtÅtkZesaykHk=vkos'k sadsfudk;dhfo|qrfLFkfrtÅtkZesaâkl
1 mV 2 + 1 2m V 2 =
1 2q2
1 2q2
2 A 2
B 4o 𝑙 o
4o 2𝑙 o
1 1 V 2
1 q2
or ;k
mV 2 +
2m A = .
2 A 2
2
4o 𝑙 o
Solving we get the speed of charge A is VA = .
gy djus ij ge izkIr djrs gS] vkos'k Adh pky VA = .
16.49 The work done by electrostatic force on charge A while the separation between both charges changes from 𝑙0 to 2𝑙0 is.
tcnksuksavkos'k sadschpdhnwjh𝑙0 lscnydj2𝑙0 gkstkrhgS]rksvkos'kAijfLFkjfo|qrcy}kjkfd;kx;kdk;Z
gS&
q2
(A) 120 𝑙
(B*)
q2 60 𝑙
(C)
q2 40 𝑙
(D)
q2 240 𝑙
Sol. (Moderate) The work done by electrostatic force on charge A, from work energy theorem, in the given duration is
= Final kinetic energy of charge A – Initial kinetic energy of charge A = 1 mV 2 – 0
2 A
1 q2
= 6o 𝑙 o .The sign of work done is positive
dk;Z&ÅtkZizes;ls]vkos'kAijfLFkjfo|qrdh;cy}kjkfd;kx;kdk;Z]fn;sx;svUrjkyesa]
= vkos'k Adh vfUre xfrt ÅtkZ – vkos'kAdh izkjfEHkd xfrt ÅtkZ = 1 mV 2 – 0
2 A
1 q2
= 6o 𝑙 o
fd;sx;sdk;ZdkfpUg/kukRedgSA
16.50 Total work done by electrostatic force on charge A + charge B while the separation between both
charges changes from 𝑙0 to 2𝑙0 is.
tcnksuksavkos'k sadschpdhnwjh𝑙0 lscnydj2𝑙0 gkstkrhgS]rksvkos'kA+vkos'kB ijfLFkjfo|qrcy}kjkfd;k x;kdqy dk;ZgS &
q2
(A) 120 𝑙
(B)
q2 60 𝑙
(C*)
q2 40 𝑙
(D)
q2 240 𝑙
Sol. (Easy) The net work done by electrostatic force on system of two charged particle is equal to change in electrostatic potential energy of the system
1 2q2
1 2q2 1 q2
= 4o 𝑙 o
4o 2𝑙 o = 4o 𝑙 o . The sign of work done is positive
(Easy) nksvkosf'krd.kksadsfudk;ijfLFkjfo|qrdh;cy}kjkfd;kx;kusVdk;Zfudk;dhfLFkjfo|qrfLFkfrt
ÅtkZesaifjorZudscjkcjgksrkgSA
1 2q2
1 2q2 1 q2
= 4o 𝑙 o
4o 2𝑙 o = 4o 𝑙 o fd;sx;sdk;ZdkfpUg/kukRedgSA
COMPREHENSION
A charge q is divided into three equal parts and placed symmetrically on a circle of radius r. The same charge is divided into four equal parts and placed symmetrically on the same circle. The electric field intensities at the centre of the circle in two situations are zero
,d vkos'k q dks rhu leku Hkkxks esa foHkDr dj r f=kT;k ds o`Ùk ij lefer
:i esa j[k fn, tkrs gSA blh vkos'k (q) dks pkj leku Hkkxksa esa foHkDr dj blh o`Ùk ij lefer :i esa j[k fn;k tkrk gSA o`Ùk ds dsUnz ij fo|qr
{kS=k rhozrk nksuks ifjfLFkfr;kas esa 'kwU; gSA
16.51 The ratio of electric potentials at the centre in the two situations is
nksfLFkfr;ksesadsUnzijfo|qrfoHkodkvuqikrgSA
(A)
(B*) 1
1
4
(C) 3
16
(D) 9
16.52 The potential energy of the system in first situation where the charge is divided into three equal parts is
ra=kdhfLFkfrtÅtkZfLFkfresatcfdvkos'kdksrhuHkxksesafoHkDrfd;kx;kgSA
(A)
1 q2
4o r
(B)
1 q2
36o r
(C*)
1 q2
(D) 12o r
16.53 If a charge (part charge) is removed from one location in both the situations, the ratio of magnitudes of the electric field intensities at the centre is
;fnnksuksafLFkfr;kasesa,dvkos'kdksgVkfn;ktk,rksdsUnzijfo|qr{ks=krhozrkdsifjek.kdkvuqikrgSA
(A) 1 2
(B) 1
1
2
(C) 3
4
(D*) 3
Sol. Potentials at the centre dsUnz ij foHko
1 q 1 q
v1 =
4o r ; v2 = 4o r
Potential energy in situation is
fLFkfresafLFkfrtÅtkZ
1 (q/ 3)2
U1 = 3 ×
4o (
3 R) =
When one charge is removed, the field
intensity at the centre is due to the removed charge only.
tc,dvkos'kdksgVkrsgSrksdsUnzijfo|qr{ks=kdhrhozrkgVk,x;svkos'kdsdkj.kghgksxhA
1 q/ 3
E1 =
4o r 2
1 q/ 4
E =
E1 4
2 4o r 2 E2 3
COMPREHENSION
A thin ring of radius R metres is placed in x-y plane such that its centre lies on origin. The half ring in region x< 0 carries uniform linear charge density + C/m and the remaining half ring in region x> 0 carries uniform linear charge density – C/m.
R f=kT;k dh iryh oy; dks x-y ry esa bl izdkj j[kk tkrk gS fd bldk dsUnz ewyfcUnq ij jgrk gSA x< 0 {ks=k okyh vk/kh oy; ij + C/m le:i js[kh; vkos'k ?kuRo gS rFkk 'ks"k vk/kh oy; ij (x> 0 {ks=k esa) – C/m le:i js[kh; vkos'k ?kuRo gSA
y´
R
16.54 Then the electric potential (in volts) at point P whose coordinates are (0m, + 2 m) is
rksfcUnqPftldsfunsZ'k ad
R
(0m,+ 2 m)
gSijfo|qrfoHkogksxk&
(A)
1
4o 2
1
(B*) 0
(C)
4o 4
(D) cannot be determined Kkr ugh fd;k tk ldrk
Sol. Consider two small elements of ring having charges +dq and – dq symmetrically located about y-axis. The potential due to this pair at any point on y-axis is zero. The sum of potential due to all such possible pairs is zero at all points on y-axis.
Hence potential at P(0,
R ) is zero.
2
ekukfdoy;dsnksNksVsHkkxftuijvkos'k+dqvkSj–dqgS]y-v{kdsifjr%lefefr:ilsfLFkrgSAbl;qXe dsdkj.ky-v{kijfdlhfcUnqijfoHko'kwU;gksxkA,slslHkhlaHko;qXeksadsfoHkoksadk;ksxy-v{kijlHkhfcUnqvksa ij'kwU;gksxk
vr%P(0, R )ijfoHko'kwU;gSA
x´ x
y´
R
16.55 The direction of electric field at point P whose coordinates are (0m, + 2 m) is
(A*) Along positive x-direction (B) Along negative x-direction
(C) Along negative y-direction (D) None of these
fcUnqPftldsfunsZ'k ad
R
(0m,+ 2 m)
gSijfo|qr{ks=kdhfn'k gS&
(A*) /kukRedx-fn'k dsvuqfn'k (B) _.kkRedx-fn'kkds vuqfn'k
(C) _.kkRedy-fn'kkds vuqfn'k (D)mijksDresalsdksbZugha
Sol. Since all charge lies in x-y plane, hence direction of electric field at point P should be in x-y plane Also y-axis is an equipotential (zero potential) line. Hence direction of electric field at all point on y- axis should be normal to y-axis.
The direction of electric field at P should be in x-y plane and normal to y-axis. Hence direction of
electric field is along positive-x direction.
Sol. pw¡fdlHkhvkos'kx-yryesafLFkrgS]blfy,fcUnqPijfo|qr{ks=kdhfn'k x-yryesagksuhpkfg,A
y-v{klefoHko¼'kwU;foHko½js[k HkhgSAvr%y-v{kijlHkhfcUnqvksaijfo|qr{ks=kdhfn'k y-v{kdsyEcor~gksuh pkfg,A
vr%Pijfo|qr{ks=kdhfn'k x-yryesagksuhpkfg,vkSjy-v{kdsyEcor~gksuhpkfg,Avr%fo|qr{ks=kdhfn'k
/kukRedx-v{kdsvuqfn'kgksxhA
16.56 The dipole moment of the ring in C–m is
oy;dkf}/kzqovk?kw.kZC–m esagSA
(A) – (2R2) ˆi
(B) (2R2) ˆi
(C*) – (4R2) ˆi
(D) (4R2) ˆi
Sol. Consider two small elements of ring having charge +dq and –dq as shown in figure.
The pair constitutes a dipole of dipole moment. dp = dq 2R = ( Rd) 2R
The net dipole moment of system is vector sum
of dipole moments of all such pairs of elementary charges. By symmetry, the resultant dipole moment is
along negative x-direction.
/ 2
x´ x
/ 2
net dipole moment = –
= – 4R2 ˆi
(dp cos ) ˆi
/ 2
(2R2 cos d) ˆi
/ 2
y´
Sol. ekukfdoy;dsnksNksVsHkkxftuijvkos'k+dqvkSj–dqgStSlkfdfp=kesafn[kk;kx;kgSA
;g;qXe,df}/kzqocukrkgS]ftldkf}/kzqovk?kw.kZgS
dp = dq 2R = ( Rd) 2R ra=kdkifj.kehf}/kzqovk?kw.kZ,slslHkhvkos'ksads;qXeksadsf}/kzqovk?kw.ks± dklfn'k;ksxgSAlefefrls]
ifj.k ehf}/kzqovk?kw.kZ_.k Redx-v{kdsvuqfn'kgS
usVf}/kzqovk?kw.kZ=–
/ 2
(dp cos ) ˆi
/ 2
/ 2
(2R2 cos d) ˆi
/ 2
x´ x
y´
= – 4R2 ˆi
SECTION - V : MATRIX - MATCH TYPE
[k.M- V:eSfVªDl&lqesy izdkj
16.57 In each situation of column-I, some charge distributions are given with all details explained.The electrostatic potential energy and its nature is given in column -II. Then match situation in column-I with the corresponding results in column-II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the OMR.
LrEHk I esa izR;sd fLFkfr esa dqN vkos'k forj.k of.kZr dj fn;k x;k gSA LrEHk-II esa fLFkjoS|qr fLFkfrt ÅtkZ o bldh izd`fr nh xbZ gSA LrEHk -I esa nh xbZ fLFkfr;ksa dks muds laxr ifj.kkeks dks LrEHk-II esa lqesfyr dfj, rFkk vius mÙkjks dks mfpr cqYyks dks xgjk dj OMR esa nh xbZ 4 × 4 dh esfVªDl esa nhft,A
Column-I Column-II
(A) A thin shell of radius a and having
a charge – Q uniformly distributed (p) over its surface as shown
(B) A thin shell of radius 5a and having
2
a charge – Q uniformly distributed (q) over its surface and a point charge
– Q placed at its centre as shown.
(C) A solid sphere of radius a and having
a charge – Q uniformly distributed (r) throughout its volume as shown.
1 Q2
8 0 a
3 Q2
20 0 a
2 Q2
5 0 a
in magnitude
in magnitude
in magnitude
(D) A solid sphere of radius a and having a charge – Q uniformly distributed
throughout its volume. The solid sphere (s) Positive in sign is surrounded by a concentric thin
uniformly charged spherical shell of radius 2a and carrying charge –Q
as shown (t) zero
LrEHkI LrEHkII
(A) ,d irykdks'k ftldhf=kT;k arFkk
1 Q2
– Q vkos'k bldh lrg ij ,d leku (p) ifjek.kesa 8 0 a
:ilsfp=k uqlkjforfjrgS
(B) ,d iryk dks'k ftldh f=kT;k 5a o
vkos'k– Q ,d leku :i ls bldh (q)
3
20 0
Q2
a ifjek.kesa
lrg ij forfjr gS rFkk ,d – Q fcUnq vkos'kbldsdsUnzijfp=k uqlkjj[k gSA
(C) ,dBkslxksykftldhf=kT;karFkk
vkos'k – Q gS rFkk ;g ,d leku :i (r)
lsbldsvk;ruesaiwjhrjgfp=kkuqlkj forfjrgSA
2 Q2
5 0 a
ifjek.kesa
(D) ,dBkslxksykftldhf=kT;kagSrFkk–Q vkos'kblds vk;ruesa iwjhrjg ,dleku
:i ls forfjr gSA ;g Bksl xksyk pkjkas vksj ls ldsUnzh;dks'k ftldh f=kT;k 2als (s)fpUgesa/kukRed f?kjkgqvkgSrFkkblij,dleku
:ils–Qvkos'kfp=k uqlkjforfjrgSA (t)'kwU;
Ans. (A) p, s (B) q, s (C) q, s (D) s
Sol. (A) Electrostatic potential energy =
1
4 0
(Q)2
2a
Q2
= 8 0 a
1 (Q) (Q)
(Q)2
3 Q2
(B) Electrostatic potential energy =
4 5a / 2 2(5a / 2) = 20 a
0 0
1 3Q2 3 Q2
(C) Electrostatic potential energy =
4 0 5a = 20 0 a
1 3Q2 (Q)2 (Q) (Q)
27Q2
(D) Electrostatic potential energy =
4
5a
2(2a)
2a
= 80 a
0
1 (Q)2 Q2
0
(A) fLFkj oS|qr fLFkfrt ÅtkZ = 4 0
2a = 8 0 a
1 (Q) (Q)
(Q)2
3 Q2
(B) fLFkj oS|qr fLFkfrt ÅtkZ = 4
5a / 2
2(5a / 2) = 20 a
0
1 3Q2
0
3 Q2
(C) fLFkj oS|qr fLFkfrt ÅtkZ = 4 0
5a = 20 0 a
1 3Q2 (Q)2 (Q) (Q)
27Q2
(D) fLFkj oS|qr fLFkfrt ÅtkZ = 4
5a
2(2a)
2a
= 80 a
0
0
16.58 Column gives certain situations involving two thin conducting shells connected by a conducting wire via a key K. In all situations one sphere has net charge +q and other sphere has no net charge. After the key K is pressed, column gives some resulting effect. Match the figures in Column with the statements in Column
and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS. LrEHkesanksiryspkyddks'ktksdqathKdhlgk;rkls,dpkydrkj}kjktqM+sgq,gStSlhdqNfLFkfr;kalfEefyr gSAlHkhfLFkfr;ksaesa,dxksysijifj.k ehvkos'k+qrFk nwljsijifj.k ehvkos'kughgSAdqathKdsnckusdsckn] LrEHk esaifj.k ehizHk ofn,gSALrEHkdsfp=k sadksLrEHk dsdFkuksadslkFklqesfyrdjsrFk vkidsmRrjdks ORS esa 4 × 4 esfVªDl esa mfpr cqycqys dks xgjk dj ds fn[kk,A
Column Ι Column ΙΙ
+q K
initially no net charge
(A)
shell
shell
(p) charge flows through connecting wire
tqM+hgqbZrkjesalsvkos'kizokggksrkgSA
+q initially no
(B)
K net charge
(q) Potential energy of system of spheres decreases.
shell
shell
xksyksdsra=kdhfLFkfrtÅtkZdegksrhgSA
initially no net charge
(C) (r) Noheatisproduced.m"ek mRiUu ugh gksrh gSA
shell
+q
(D) (s) The sphere has no charge after equilibrium is reached.
shell
Ans. (A) p, q (B) p, q (C) p, q, s (D) r, s, t
lkE;koLFk izkIrdjusdsi'pkr~xksysijvkos'kughgSA
(t) charge does not flows through connecting wire
tqM+hgqbZrkjesalsvkos'kizokgughagksrkgSA
Sol. In situation A, B and C, shells and are not at same potential. Hence charge shall flow from sphere to sphere till both acquire same potential.
If charge flows, the potential energy of system decreases and heat is produced.
In situations A and B charges shall divide in some fixed ratio , but in situation C complete charge shall be transferred to shell II for potential of shell and to be same.
(A) p, q , (B) p, q , (C) p, q, s
In situation D both the shells are at same potential, hence no charge flows through connecting wire.
(D) r, s
Sol. A,BvkSjCfLFkfr;ksaesa]dks'kdks'k lekufoHkoijughagSAblfy;svkos'kxksyklsxksyk esarcrdizokfgr gksrkgS]tcrdfdnksuksalekufoHkoizkIrdjrsgSaA
;fnvkos'kizokfgrgksrkgS]ra=kdhfLFkfrtÅtkZ?kVtkrhgSvkSjÅ"ekmRiUugksrhgSA AvkSjBfLFkfr;ksaesa]vkos'k,dfuf'prvuqikresac¡VtkrsgSa]ysfdufLFkfrCesadks'kodks'k dkfoHkoleku gksusdsfy,iwjkvkos'kdks'k esaizokfgrgkstk;sxkA
(A) p, q , (B) p, q , (C) p, q, s
fLFkfrDesanksuksadks'klekufoHkoijgSa]vr%vkos'ktksM+usokysrkjesaizokfgrughagksrkgSA
(D) r, s
SECTION - VI : INTEGER TYPE
[k.M - VI : iw.kkZad izdkj
SHORT SUBJECTIVE
y?kqÙkjkRediz'u
16.59 A positive charge +Q is fixed at a point A. Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in the figure. The point B is at large distance from A and at distance ‘d’ from the line AC. The initial velocity is parallel to the line AC. The point C is at very large distance
from A. Find the minimum distance (in meter) of +q from +Q during the motion. Take Qq = 4 mu2d
and
d (
1) meter.
,d /kukRed vkos'k +Q fcUnq A ij tM+or~ gSA ,d vU; /kukosf'kr d.k ftldk nzO;eku m rFkk vkos'k +q gS] fcUnq B ls u osx ls fp=kkuqlkj iz{ksfir fd;k tkrk gS fcUnq B fcUnq A ls cgqr vf/kd nwjh ij gS rFkk js[kk AC ls d nwjh ij gSA izkjfEHkd osx js[kk AC ds lekUrj gSA fcUnq C fcUnq A ls cgqr vf/kd nwjh ij fLFkr gSA xfr ds nkSjku +q dh +Q ls U;wure nwjh
(ehVj esa) Kkr djksA Qq = 40
mu2d rFkk
d (
1) ehVj gSA
Ans. 1
Sol. The path of the particle will be as shown in the figure. At the point of minimum distance (D) the velocity of the particle will be to its position vector w.r. to +Q.
Now by conservation of energy :- d.kdkiFkfp=kesaiznf'kZrgSAU;wurenwjhokysfcUnq(D)ijd.kdkosx]+QlsbldsfLFkfrlfn'kdslkis{kyEcor~ gkstkrkgSA
vcÅtkZlaj{k.kls
1 1 KQq
2 mu2 + 0 = 2
mv2 +
rmin
......(1)
Torque on q about Q is zero hence angular momentum about Q will be conserved
Q ds lkis{k q ij cyk?kw.kZ 'kwU; gS vr% Q ds lkis{k dks.kh; laosx lajf{kr jgsxk
m v r = m ud (2)
1 1 ud 2 KQq
by (2) in (1) (lehdj.k (2) ls (1) esa) 2 mu2 = 2 m r r
1 1 d
mu2d
min
min
mu2
2
min
rmin
{ KQq= mu2d(givenfn;k gS) }
2
min
– 2rmin
d – d2 = 0 r
min = 2
= d (1 ± )
distancecannotbe negativenwjh _.kkRed ugh gks ldrh gSA
r = d(1 + ) Ans.
16.60 Consider a cube of side a = 0.1 m placed such that its six faces are given by equations x = 0, x = +a,
y = 0, y = +a, z = 0 and z = +a, placed in electric field given by crossing out of the cube in the unit of 10–4 N m2/C.
→ x2ˆi y ˆj N/C. Find the electric flux
Hkqtk a = 0.1 m dk ,d ?ku yhft, ftlds N% i`"B tks lehdj.k x = 0, x = +a, y = 0, y = +a, z = 0 vkSjz =
+a }kjk fn;s tkrs gSa] fo|qr {ks=k → x2ˆi y ˆj N/Cesa fLFkr gSA ?ku ls fudyus okys fo|qr ¶yDl dks
10–4 N m2/C ek=kd ds :i esa Kkr dhft,A
Ans. 11
Sol. (Moderate) Flux through ABCD. ABCD ls gksdj fo|qr ¶yDl
→ →
1 = E A
= ( x2ˆi y ˆj ). ( a2 ˆi )
= 0 as x = 0
Flux through EFGH EFGH ls gksdj fo|qr ¶yDl
= ( x2ˆi y ˆj ). ( a2 ˆi ) y
= x2.a2 = a4 = 1.0 × 10–4 Nm2/C
Flux through BCGF BCGF ls gksdj fo|qr ¶yDl x
= ( x2ˆi y ˆj ). ( a2 ˆj )
= a3 = 1.0 × 10–3 Nm2/C
Flux through EADH EADH ls gksdj fo|qr ¶yDl
= ( x2ˆi y ˆj ). ( a2 ˆj ) = 0 as y = 0
Flux through ABFE ABFE ls gksdj fo|qr ¶yDl
= ( x2ˆi y ˆj ). ( a2 kˆ ) = 0
Flux through CDHG CDHG ls gksdj fo|qr ¶yDl
6 = 0
Net flux (usV ¶yDl) = (1.0 × 10–4 + 1.0 × 10–3) N-m2/C = 11 × 10–4 N-m2/C
R
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