Chapter # 31
1.
Sol.
Capacitors
[1]
Objective - I
A capacitor of capacitance C is charged to a pontential V. The flux of the electric field through a closed surface enclosing the capacitor is C /kkfjrk dk ,d la/kkfj=k V foHko rd vkosf'kr fd;k x;k gSA la/kkfj=k dks ifjc) djus okys can i`"B ls ikfjr oS|qr ¶yDl gSA (A) CV/o (B) 2CV/o (C) CV/2o (D*) zero D
Q=cv
S
C
Q + vS :- Closed surface enclosing the capacitor. no. of electricfied lines enter in the closed path = no. of electric field lines exist in the closed Path.
0 So we can say that the flux of the el3ectric field through a closed surface enclosing the capacitor is zero. 2.
Sol.
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combinate will be C /kkfjrk izR;sd ds nks la/kkfj=k] ftudh Hkatu oksYVrk V gS] Js.khØe esa tksM+s tkrs gSA la;kstu dh /kkfjrk ,oa Hkatu oksYVrk gksxh (A) 2 C and 2 V (B) 2/C and 2/V (C*) 2 C and 2/V (D) 2/C and 2 V D +v+vC
C
CC C CC 2 Voltage across Ceq. = v + v = 2v Ceq.
3.
Sol.
4.
Sol.
If the capacitors in the previous question are joined in paralle, the capacitance and the breakdown voltage of the combination will be ;fn iwoZ iz'u esa la/kkfj=k lekUrj Øe esa tksM+s tk, rks la;kstu dh /kkfjrk ,oa Hkatu oksYVrk gksxh (A) 2 C and 2 V (B*) C and 2 V +v(C) 2 C and V (D) C and V C C Ceq. = C + C = 2C +vVoltage across Ceq. = v C The equivalent capacitance of the combination shown in figure is fp=k esa n'kkZ, la;kstu dh rqY; /kkfjrk gS -
B
(A) C
(B*) 2 C
(C) C/2
(D) none of these
C1=C C2=C
C3=C
Here voltage cross ‘C3’ is same across terminal charge flow through C3 = 0
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Chapter # 31
Capacitors
[2]
C
5.
Sol.
Ceq = C + C = 2C
C
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plated will ,d foyfxr la/kkfj=k dh IysVksa ds e/; ,d ijkoS|qr inkFkZ dh ifV~Vdk j[kh tkrh gSA IysVksa ds e/; cy (A) increase (B) decrease (C*) remain unchanged (D) become zero (A) c<+sxk (B) ?kVsxk (C*) vifjofrZr jgsxk (D) 'kwU; gks tk,xk C Force between capacitor plates :q = A,
E
2 0
+ + + +
F = qE
F A.
2 A 2 0 2 0
F-
After inserted dielectric slap, charge on the other plate is remain same & the fore between the plates will remain uncharged. 6.
Sol.
The energy density in the electric field created by a point charge falls off with the distance from the point charge as ,d fcUnq vkos'k }kjk mRiUu fo|qr {ks=k esa ÅtkZ ?kuRo fcUnq vkos'k ls nwjh ds lkFk ?kVrk gS (A) 1/r (B) 1/r2 (C) 1/r3 (D*) 1/r4 D 2
Energy density
1 1 Q 0 E2 0 4 r 2 2 2 0
2 1 0 Q 2 162 0 r 4
Energy density 7.
1 r4
A parallel-plate capacitor has plates of unequal of unequal area. The large plate is connected to the positive terminal of the battery and the smaller plate to its neagtive terminal. Let Q+ and Q– be the charges apperaing on the positive and the negative plates respectively.
,d lekUrj iV~V la/kkfj=k dh IysVksa dk {ks=kQy leku ugha gSA cM+h IysV dks cSVjh ds /ku VfeZuy rFkk NksVh IysV dks nwljs _.k VfeZuy ls tksM+k tkrk gSA ekuk Q+ o Q– Øe'k% /ku o _.k IysV ij vkos'k gS rks -
Sol.
(A) Q+ > Q–. (B*) Q+ = Q–. (C) Q+ < Q–. (D) The infromation is not sufficent to decide the relation between Q+ and Q-. Q+ rFkk Q– ds chp laca/k Kkr djus ds fy, lwpuk vi;kZIr gSA +Q -Q B Charge not depends upon the Area of plates. So Q+ = Q-
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Chapter # 31 Capacitors [3] 8. A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates (fig.). The capacitance now becomes C /kkfjrk ds lekUrj iV~V la/kkfj=k dh IysVksa ds e/; ,d iryh /kkrq dh IysV P bl rjg izosf'kr djkbZ tkrh gSA fd blds fdukjs nksuksa IysVksa dks Li'kZ djsA /kkfjrk vc gks tkrh gS -
Sol.
D
(A) C/2
(B) 2 C
(D*)
(C) 0
d Thin metal plate edges touch the two plates that causes the ‘d’ is reduces to zero.
C 9.
0 A d
Fig. shows two capacitors connected in series and joined to a bettery. The graph shows the variation in pontential as one moves from left to right on the branch containing the capacitors -
fp=kkuqlkj nks la/kkfj=k dks Js.khØe esa tksM+dj ,d cSVjh ls tksMk+ tkrk gSA la/kkfj=kksa okyh 'kk[k esa cka;s ls nka;s pyus ij foHko esa ifjorZu xzkQ }kjk n'kkZ;k x;k gS (A) C1 > C2.
(A)
Sol.
C
(B) C1 = C2.
(C*) C1 < C2.
(B)
(D) The information is not sufficient to decide the relation between C1 and C2. v v (D) C1 o C2 ds e/; laca/k Kkr djus ds fy, lwpuk vi;kZIr gSA
v x C2 > C1 E
10.
Sol.
C1
C2
x
Two metal plates having charges Q, - Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will (A*) increase (B) decrease (C) remain the same (D) become zero nks /kkfRod IysVsa] ftu ij vkos'k Q, o - Q gS] ,d nwljs ls dqN nwjh ij vkeus lkeus gSA bUgsa rsy ds ,d VSad esa Mqcks;k x;k gSA ;fn rsy ckgj fudky fn;k tkrk gS] rks IysVksa ds e/; fo|qr {ks=k (A*) c<+sxk (B) ?kVsxk (C) ogh jgsxk (D) 'kwU; gks tk,xk A
E
v Q d 2 0 A
Eoil
Q 2 0 KA
E > Eoil If oil is pumpedout, the electrif field between the plates will increased.s
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Chapter # 31 Capacitors [4] 11. Two spheres of capacitances C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy. C1 o C2 /kkfjrk ds nks /kkfRod xksyksa ij dqN vkos'k gSA mUgsa ,d nwljs ls laifdZr dj fQj vyx dj fn;k tkrk gSA mu ij vafre vkos'k Q1 o Q2 fuEu laca/k dks larq"V djsaxs (A) Sol.
Q1 C 1 Q 2 C2
(B*)
Q1 C 1 Q 2 C2
(C)
Q1 C 1 Q 2 C2
(D)
Q1 C 2 Q 2 C1
B Charge flow unit 1 potential diference accross the capacitor will be same. Then final charge is Q, at C1 & Q2 at C2. at capacitor “C1” Q1 = C1 Q v
Q1 C1
....(i)
at capacitor “C2” Q2 = C2 v v
Q2 C2
....(ii)
From eq. (i) & (ii) we get Q1 Q2 C1 C2
12.
Sol.
Q1 C1 Q1 C2
Three capacitors of capacitances 2F each are available. The minimum and maximum capacitances, which may be obtained are rhu la/kkfj=k ftuesa izR;sd dh /kkfjrk 2F gSA mlls izkIr U;wure o vf/kdre /kkfjrk gksxh (A) 6F, 18F (B) 3F, 12F (C) 2F, 12F (D*) 2F, 18F D for minimum capacitances :Ceq Capacitor cannect in series =
1 1 1 1 Ceq 6 6 6
Ceq 2f
for maximum capacitances :Capacitor cannect in parallel
=
Ceq
Ceq. = 6 + 6 + 6 = 18f
Objective - II 1.
The capacitance of a capacitor does not depend on (A) the shape of the plates (B) the size of the plates (C*) the charges on the plates (D) the separation between the plates. ,d la/kkfj=k dh /kkfjrk fuHkZj ugha djrh gS (A) IysVksa dh vkÑfr ij (B) IysVksa ds vkdkj ij (C*) IysVksa ds vkos'k ij (D) IysVksa ds e/; nwjh ij
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Chapter # 31 Sol. C
Capacitors
[5]
0 A d Capacitance of capacitor does not depends on the charges at the plates. C
2.
Sol.
A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantites will remain the same ? (A) The electric field in the capacitor (B*) the charge on the capacitor (C) the potential differencr between the plates (D) the stored energy in the capacitor
,d foyfxr vkosf'kr la/kkfj=k dh IysVksa ds e/; ,d ijkoS|qr ifV~Vdk j[kh tkrh gSA fuEu esa ls dkSulh jkf'k;k¡ ugha cnysaxh& (A) la/kkfj=k esa fo|qr {ks=k (B*) la/kkfj=k ij vkos'k (C) IysVksa ds e/; foHkokUrj (D) la/kkfj=k esa laxfgr ÅtkZ
B a dielectric slap is inserted between the plates of an isolated charged capacitor. The charge on the capacitor remain same.
C
K 0 A d
K - Dielectic constant of the slab
E
Q 2 0 KA
C - Capacitance E - Electric field
U
3.
Sol.
4.
Sol.
1 1 0 E2 0 2 2
L
Q 2 0KA
U - Energy storedin capacitor
A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q’. (A) Q’ may be larger than Q. (B) Q’ must be large than Q. (C) Q’ must be equal than Q. (D*) Q’ must be smaller than Q. ,d la/kkfj=k dh IysVksa ds e/; ,d ijkoS|qr ifV~Vdk izosf'kdk djkbZ tkrh gSA la/kkfj=k ij vkos'k Q gS rFkk ijkoS|qr ifV~Vdk ds izR;sd i`"B ij izsfjr vkos'k dk ifjek.k Q’ gS (A) Q’ , Q ls T;knk gks ldrk gSA (B) Q’ , Q ls T;knk gksuk pkfg,A (C) Q’ , Q ds cjkcj gksuk pkfg,A (D*) Q’ , Q ls de gksuk pkfg,A D Q’ must be smaller than Q. Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now, (A*) the facing surfaces of the capacitor have equal and opposite charges. (B) the two plates of the capacitor have equal and opposite charges. (C*) the battery supplies equal and oppostie charges to the two plates. (D*) the outer surface of the plates have equal charges. ,d lekUrj iV~V la/kkfj=k dh izR;sd IysV ij q vkos'k gSA la/kkfj=k dks vc ,d cSVjh ls tksM+k tkrk gSA vc (A*) la/kkfj=k ds vkeus&lkeus ds i`"Bksa ij cjkcj ,oa foifjr vkos'k gSA (B) la/kkfj=k dks nksuksa IysVksa ij cjkcj ,oa foifjr vkos'k gSA (C*) cSVjh nksuksa IysVksa dks cjkcj ,oa foifjr vkos'k nsrh gSA (D*) IysVksa ds ckgjh i`"Bksa ij leku vkos'k gSA ACD q0 +q -q
2
The capacitor is now connected to a battery. Now the facing surfaces of the capacitor have equal and opposite sharge The battery supplies equal and opposite charges to the two plates. The outer surface of the plates have equal charges.
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2
v
Chapter # 31 Capacitors [6] 5. The separation between the plates of a charged parallelplate capacitor is increased. Which of the following quantities will charge ? (A) charge on the capacitor (B*) potential difference across the capacitor (C*) energy of the capacitor (D) energy density between the plates. ,d vkosf'kr lekUrj iV~V la/kkfj=k dh IysVksa ds e/; nwjh c<+kbZ tkrh gSA buesa ls dkSulh jkf'k;k¡ ifjofrZr gksxh (A) la/kkfj=k ij vkos'k (B*) la/kkfj=k ij foHkokUrj (C*) la/kkfj=k dh ÅtkZ (D) IysVksa ds e/; ÅtkZ ?kuRo Sol. BC d
C
0 A d
d’ > d
d'
C'
0 A d'
C’ < C Charge on the capacitor will remain same. Q = Cv = C’ v’ C > C’, So V’ > V V + V’ are the potential difference across the capacitor. Energy ‘V’ =
Q2 2c
New energy u'
6.
Sol.
C > C’ ,
Q2 2c ' So U’ > U
A parallel-plate capacitor is connected to a battery. A metal sheet of negligable thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor. (A) The battery will supply more charge (B) The capacitance will increase. (C) The potnetial difference between the plates will increase (D*) Equal and opposite charges will appear on the two faces of the metal plate.
,d lekUrj iV~V la/kkfj=k ,d cSVjh ls tksMk+ tkrk gSa IysVksa ds e/; ,d ux.; eksVkbZ dh /kkfRod iV~Vh j[kh tkrh gSA iV~Vh la/kkfj=k dh IysVksa ds lekUrj jgrh gS (A) cSVjh vf/kd vkos'k nsxhA (B) /kkfjrk c<+ tk,xhA (C) IysVksa ds e/; foHkokUrj c<+ tk,xkA (D*) /kkrq dh IysV dh nksuksa lrgksa ij cjkcj ,oa foifjr vkos'k vk tk,xkA D
-q q +Q -Q Metal Plate
+ v Equal and opposite charge will appear on the two faces of the metal plate. 7.
Following operations can be performed on a capacitor: X - connect the capacitor to a battery of emf w. Y - disconnect the battery Z - reconnect the battery with polarity reversed W - insert a dilectric slab in the capacitor (A) In XYZ (peroform X, then Y, then Z) the stored electric energy remains unchanged and no termal energy is developed. (B*) The charge appearing on the capacitor is greater after the action XWY than after the action XYW (C*) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW. (D*) The electric field in the capacitor after the action XW is the same as that after WX.
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Chapter # 31
Sol.
Capacitors
[7]
,d la/kkfj=k ij fuEu lafØ;k,sa dh tk ldrh gSa : X - la/kkfj=k dks fo|qr okgd cy dh cSVjh ls tksM+ukA Y - cSVjh dks gVkuk Z - cSVjh dks foijhr /kzqo.krk ls iqu% tksM+ukA W - la/kkfj=k esa ,d ijkoS|qr ifV~Vdk ?kqlkukA (A) XYZ esa (igys X djsa , fQj Y, fQj Z) laxzfgr oS|qr ÅtkZ vifjofrZr jgrh gS ,oa dksbZ rkih; ÅtkZ mRiUu ugha gksrhA (B*) fØ;k XYW ds i'pkr~ la/kkfj=k ij vk;k vkos'k] fØ;k xyw ds i'pkr~ vk;s vkos'k ls T;knk gSA (C*) fØ;k XWY ds i'pkr~ la/kkfj=k esa laxzfgr ÅtkZ] fØ;k xyw ds i'pkr~ laxzfgr ÅtkZ ls T;knk gSA (D*) fØ;k XW ds i'pkr~ la/kkfj=k esa oS|qr {ks=k] fØ;k WX ds i'pkr~ oS|qr {ks=k ds cjkcj gSA BCD A xyz
C + -
C
Ui
1 c 2 2
Uf
1 c 2 2
+ work doen by the battery = 2c 2 Heat = W.D. by the battery - (Uf - Ui) = 2c 2 Thermal Energy B xwy
K
C
x
w
y C'=CK
+ -
+ -
xyw
x
C
y
w
K C'=CK
+ Q > Q1 The charge appearing on the capacitor is greater after the action xwy than after the action xyw. C
w
K
x
y
C'=CK
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Chapter # 31
Capacitors
[8]
Q2 C2 2 1 C 2 2C ' 2CK 2 K The electric Energy stored in the capacitor is greater after the action wxy then after the action xyw. Electric Energy = E
D xw
K x
w
C
C'=CK
+ -
+ -
Q C K C Electric field = 2A K 2A K 2A 0 0 0 wx
w
O
K O
C'=CK
x
C'
-C' Q = C'
=CK
+ -
Q C K C Electric field = 2A K 2A K 2A 0 0 0
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