The potential energy of charged parallel plate capacitor is $v_{0}$. If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be
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- $\frac{v_{0{K}$
- $v_{0}K^{2}$
- $\frac{v_{0{K^{2$
- $v_{0}^{2}$
The Correct Option is A
Solution and Explanation
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A circuit consisting of a capacitor C, a resistor of resistance R and an ideal battery of emf V, as shown in figure is known as RC series circuit.
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