A slab of dielectric constant \(K\) has the same crosssectional area as the plates of a parallel plate capacitor and thickness \(\frac{3}{4}\,d\), where \(d\) is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be.(Given \(C _{0}=\) capacitance of capacitor with air as medium between plates.)

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Answer & Solution

Correct Answer

(A) \(\frac{4 KC _{0}}{3+ K }\)

Detailed explanation

\(x + y +\frac{3 d }{4}= d\)

\(x + y =\frac{ d }{4}\)

\(\frac{ A \epsilon_{0}}{ d }= C _{0}\)

\(\Delta V = Ex +\frac{ E }{ k } \times \frac{3 d }{4}+ Ey\)

\(=\frac{3 Ed }{4 k }+ E ( x + y )\)

\(\Delta V = E \left[\frac{3 d }{4 k }+\frac{ d }{4}\right]\)

\(\Delta V =\frac{\sigma}{\epsilon_{0}}\left[\frac{3 d + dk }{4 k }\right]=\frac{ Qd }{ A \epsilon_{0}}\left[\frac{3+ k }{4 k }\right]\)

\(\frac{ Q }{\Delta V }= C =\frac{ A \epsilon_{0}}{ d }\left[\frac{4 k }{3+ k }\right]=\frac{4 k C _{0}}{ k +3}\)

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