Electrostatics Question 13

Question: Separation between the plates of a parallel plate capacitor is $ d $ and the area of each plate is $ A $ . When a slab of material of dielectric constant$ k $ and thickness $ t(t<d) $ is introduced between the plates, its capacitance becomes [MP PMT 1989]

Options:

A) $ \frac{{\varepsilon _{0}}A}{d+t( 1-\frac{1}{k} )} $

B) $ \frac{{\varepsilon _{0}}A}{d+t( 1+\frac{1}{k} )} $

C) $ \frac{{\varepsilon _{0}}A}{d-t( 1-\frac{1}{k} )} $

D) $ \frac{{\varepsilon _{0}}A}{d-t( 1+\frac{1}{k} )} $

Show Answer

Answer:

Correct Answer: C

Solution:

Potential difference between the plates $ V=V _{air}+V _{medium} $

$ =\frac{\sigma }{{\varepsilon _{0}}}\times (d-t)+\frac{\sigma }{K{\varepsilon _{0}}}\times t $

therefore $ V=\frac{\sigma }{{\varepsilon _{0}}}(d-t+\frac{t}{K}) $

$ =\frac{Q}{A{\varepsilon _{0}}}(d-t+\frac{t}{K}) $

Hence capacitance $ C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A{\varepsilon _{0}}}(d-t+\frac{t}{K})} $

$ =\frac{{\varepsilon _{0}}A}{(d-t+\frac{t}{k})}=\frac{{\varepsilon _{0}}A}{d-t( 1-\frac{1}{k} )} $



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