The figure shows a parallel plate capacitor with charge Q 1 and plate area A- Capacitor is filled with two dielectrics k 1 and k 2 as shown-A- Ratio of energy density in first dielectric to second dielectric is 5-3B- Ratio of energy stored in the two dielectrics is 1- 1C- Surface charge at interface of the two dielectrics is 2 Q-15D- Surface charge at interface of the two dielectrics is 0 -

Q_1 = Q(1 1/k_1) and Q_2 = Q(1 1/k_2) So, Q_1 Q_2 = 2Q/15 Energy density in capacitor is = Energy/Volume = Q^2/2C(Volume) = Q^2/2 ϵ_o A^2k Hence, Energy ...

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Byju's Answer

Standard XII

Physics

New Capacitance in Dielectric

The figure sh...

Question

The figure shows a parallel plate capacitor with charge $Q_{1}$ and plate area A. Capacitor is filled with two dielectrics $k_{1}$ and $k_{2}$ as shown.

Ratio of energy density in first dielectric to second dielectric is

\frac{5}{3}

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Ratio of energy stored in the two dielectrics is 1 : 1

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Surface charge at interface of the two dielectrics is

\frac{2 Q}{15}

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Surface charge at interface of the two dielectrics is 0.

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Solution

The correct options are
A Ratio of energy density in first dielectric to second dielectric is $\frac{5}{3}$
C Surface charge at interface of the two dielectrics is $\frac{2 Q}{15}$

$Q_{1} = Q (1 - \frac{1}{k_{1}})$ and $Q_{2} = Q (1 - \frac{1}{k_{2}})$
So, $Q_{1} - Q_{2} = - \frac{2 Q}{15}$

Energy density in capacitor is = $\frac{E n e r g y}{V o l u m e} = \frac{Q^{2}}{2 C (V o l u m e)} = \frac{Q^{2}}{2 ϵ_{o} A^{2} k}$
Hence, $E n e r g y \propto \frac{1}{k}$
which means that the ratio of energy densities is $\frac{5}{3}$ .

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The figure shows a parallel plate capacitor with charge Q1 and plate area A. Capacitor is filled with two dielectrics k1 and k2 as shown.

The figure shows a parallel plate capacitor with charge $Q_{1}$ and plate area A. Capacitor is filled with two dielectrics $k_{1}$ and $k_{2}$ as shown.