The potential energy of a charged parallel plate capacitor is $${U_9}$$.If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be (battery disconnected)

$$\frac{{{U_9}}}{K}$$

$${U_9}{k^2}$$

$$\frac{{{U_9}}}{{{k^2}}}$$

$${U_9}$$

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Correct option is A. $$\frac{{{U_9}}}{K}$$

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