A dielectric of dielectric constant $K$ is introduced such that half of its area of a capacitor of capacitance $C$ is occupied by it. The new capacity is

Question

Dr. Abubacker · Accepted Answer

(c) The dielectric is introduced such that, half of its area is occupied by
It.
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In the given case the two capacitors are in parallel.
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = C_{1} + C_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C_{1}\  = \ \frac{A\varepsilon_{0}}{2d}$
And $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C_{2} = \ \frac{KA\varepsilon_{0}}{2d}$
Thus, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \ \frac{A\varepsilon_{0}}{2d} + \frac{KA\varepsilon_{0}}{2d}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{C}{2}(1 + K)$

Q. A dielectric of dielectric constant $K$ is introduced such that half of its area of a capacitor of capacitance $C$ is occupied by it. The new capacity is

Explanation

Reviewed by Dr. Abubacker

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