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CBSE Class 12 Physics 2014 Delhi Set 1 Paper

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Question : 24 of 30
Marks: +1, -0
Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is EE. Find the amount of work done in moving a charge qq over a closed rectangular loop abcda.
OR
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area AA and plate separation dd.
(b) Two charged spherical conductors of radii R1R_1 and R2R_2 when connected by a conducting wire acquire charge q1q_1 and q2q_2 respectively. Find the ratio of their surface charge densities in terms of their radii.
Solution:  
(b) W  =  Force  ×  Displacement  W\;=\;\text{Force}\;\times\;\text{Displacement}\;
F  =qEF\;=q E
As displacement =0=0, so work done is also zero.
(a)
E  =  σε0  =  QAε0E\;=\;\frac{\sigma}{\varepsilon_0}\;=\;\frac{Q}{A\varepsilon_0}
∴  V=Ed  =  QdAε0\therefore\;V=E d\;=\;\frac{Qd}{A\varepsilon_0}
Capacitance,  C=QV=ε0Ad\text{Capacitance,}\;C=\frac{Q}{V}=\frac{\varepsilon_0 A}{d}
(b) When the two charged spherical conductors are connected by a conducting wire they acquire the same potential.
i.e., Kq1R1  =  Kq2R2\frac{K q_1}{R_1}\;=\;\frac{K q_2}{R_2}
⇒  q1q2=R1R2\Rightarrow\;\frac{q_1}{q_2}=\frac{R_1}{R_2}
Hence, ratio of surface charge densities,
σ1σ2=q14πR12q24πR22\frac{\sigma_1}{\sigma_2}=\frac{\frac{q_1}{4\pi R_1^2}}{\frac{q_2}{4\pi R_2^2}}
σ1σ2=q1R22q2R12\frac{\sigma_1}{\sigma_2}=\frac{q_1 R_2^2}{q_2 R_1^2}
σ1σ2=R1R2×R22R12=R2R1\frac{\sigma_1}{\sigma_2}=\frac{R_1}{R_2}\times\frac{R_2^2}{R_1^2}=\frac{R_2}{R_1}
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