Electrostatics Question 651
Question: A parallel plate capacitor has capacitance C when no dielectric between the plates. Now a slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor . The new capacitance will be
Options:
A) $ (K+1)\frac{C}{4} $
B) $ (K+2)\frac{C}{4} $
C) $ (K+3)\frac{C}{4} $
D) $ \frac{KC}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We can observe that system is equivalent to two capacitors
$ C _{1} $ and $ C _{2} $ in parallel. Capacitance of original capacitor without dipole.
$ C _{1}=\frac{{\varepsilon _{0}}}{d}( \frac{3A}{4} )=\frac{3{\varepsilon _{0}}A}{4d} $
With medium, $ C _{2}=\frac{{\varepsilon _{0}}E}{d}( \frac{A}{4} )=\frac{{\varepsilon _{0}}AK}{4d} $
$ C’=C _{1}+C _{2} $ Or $ C’=\frac{3{\varepsilon _{0}}A}{4d}+\frac{{\varepsilon _{0}}AK}{4d}=\frac{{\varepsilon _{0}}A}{d}[ \frac{3}{4}+\frac{K}{4} ] $ Or $ C’=\frac{C}{4}(K+3) $
$ [ \therefore C=\frac{A{\varepsilon _{0}}}{d} ] $
BETA
