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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 26 of 35
Marks: +1, -0
SECTION - C
(a) Two charged conducting spheres of radii aa and bb are connected to each other by a wire. Find the ratio of the electric fields at their surfaces.
OR
(b) A parallel plate capacitor (A) of capacitance CC is charged by a battery to voltage VV. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C2 C is connected across AA. Find the ratio of
(i) final charges on AA and BB.
(ii) total electrostatic energy stored in A and B finally and that stored in A initially.
(a) For sphere with radius a, Potential =V= V
(since, both the charged spheres are connected by a wire)
 Charge =Qa\text{ Charge } = Q_a
 Capacitance =Ca\text{ Capacitance } = C_a
 Electric field, =Ea=Qa4πε0a2\text{ Electric field, } = E_a = \frac{Q_a}{4 \pi \varepsilon_0 a^2}
For sphere with radius bb,
Potential =V= V
(since both the charged spheres are connected by a wire)
 Charge =Qb\text{ Charge } = Q_b
 Capacitance =Cb\text{ Capacitance } = C_b
 Electric field =Eb\text{ Electric field } = E_b
=Qb4πε0b2= \frac{Q_b}{4 \pi \varepsilon_0 b^2}
 Now, EaEb=QaQb×b2a2\text{ Now, } \frac{E_a}{E_b} = \frac{Q_a}{Q_b} \times \frac{b^2}{a^2} ......(1)
QaQb=CaVCbV\frac{Q_a}{Q_b} = \frac{C_a V}{C_b V}
 Or, QaQb=CaCb\text{ Or, } \frac{Q_a}{Q_b} = \frac{C_a}{C_b}
 Or, QaQb=ab\text{ Or, } \frac{Q_a}{Q_b} = \frac{a}{b} .......(2)
(since, CdCb=ab\frac{C_d}{C_b} = \frac{a}{b} )
Putting in equation (1)
EaEb=(ab)×(b2a2)\frac{E_a}{E_b} = \left( \frac{a}{b} \right) \times \left( \frac{b^2}{a^2} \right)
∴EaEb=ba\therefore \frac{E_a}{E_b} = \frac{b}{a}
OR
(b) (i) Initially,
Charge on A capacitor =CV= CV
Charge on BB capacitor =0= 0
After connecting A with B,
 Final potential =V′= Total Charge  total capacitance \text{ Final potential } = V' = \frac{\text{ Total Charge }}{\text{ total capacitance }}
=Q+0C+2C=Q3C= \frac{Q+0}{C+2C} = \frac{Q}{3C}
Final charge on A capacitor after redistribution
=CV′=CQ3C=Q3= C V' = \frac{C Q}{3 C} = \frac{Q}{3}
Final charge on B capacitor after redistribution
=2CV′=2CQ3C=2Q3= 2 C V' = \frac{2 C Q}{3 C} = \frac{2 Q}{3}
So, the ratio of charges =Q/32Q/3=1:2= \frac{Q/3}{2Q/3} = 1:2
(ii) Initially energy stored in A=12(CV)2A = \frac{1}{2} (CV)^2
Finally energy stored in A=12CV2A = \frac{1}{2} C V^2
=12×C×(Q3C)2= \frac{1}{2} \times C \times \left( \frac{Q}{3C} \right)^2
=12×C×(V3)2= \frac{1}{2} \times C \times \left( \frac{V}{3} \right)^2
=CV218= \frac{C V^2}{18}
Finally energy stored in B=122(CV)2B = \frac{1}{2} 2 (CV)^2
=12×2C×(Q3C)2= \frac{1}{2} \times 2 C \times \left( \frac{Q}{3C} \right)^2
=12×2C×(V3)2= \frac{1}{2} \times 2 C \times \left( \frac{V}{3} \right)^2
=CV29= \frac{C V^2}{9}
Total energy stored in A and B
=CV218+CV29= \frac{C V^2}{18} + \frac{C V^2}{9}
=CV26= \frac{C V^2}{6}
So, the required ratio = Find energy stored in A and B Initial energy stored in A= \frac{\text{ Find energy stored in } A \text{ and } B}{\text{ Initial energy stored in } A}
=CV26CV22= \frac{ \frac{C V^2}{6} }{ \frac{C V^2}{2} }
=13= \frac{1}{3}
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