Two identical parallel plate capacitor are placed in series and connected to a constant voltage sour

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1. Electric Charges and Fields

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Two identical parallel plate capacitor are placed in series and connected to a constant voltage source of $V_0\, volt$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$, the potential difference between the plates of the other capacitor will change to

A

$\left( {\frac{{K + 1}}{K}} \right){V_0}$

B

$\left( {\frac{K}{{K + 1}}} \right){V_0}$

C

$\left( {\frac{{K + 1}}{2K}} \right){V_0}$

D

$\left( {\frac{2K}{{K + 1}}} \right){V_0}$

Solution

$\mathrm{V}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{1}+\mathrm{c}_{2}} \mathrm{v}_{0}=\frac{\mathrm{kc}}{\mathrm{kc}+\mathrm{c}} \mathrm{v}_{0}$

$\mathrm{V}=\left(\frac{\mathrm{k}}{\mathrm{k}+1}\right) \mathrm{V}_{0}$

Standard 12

Physics

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