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} )} $
BETA
