question_answer

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 as shown in the figure. The new capacitance will be

A) \[(K+1)\frac{C}{4}\]     

B) \[(K+2)\frac{C}{4}\]

C) \[(K+3)\frac{C}{4}\]     

D) \[\frac{KC}{4}\]

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. \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}}{d}\left( \frac{3A}{4} \right)=\frac{3{{\varepsilon }_{0}}A}{4d}\] With medium, \[{{C}_{2}}=\frac{{{\varepsilon }_{0}}E}{d}\left( \frac{A}{4} \right)=\frac{{{\varepsilon }_{0}}AK}{4d}\] \[C'={{C}_{1}}+{{C}_{2}}\] Or \[C'=\frac{3{{\varepsilon }_{0}}A}{4d}+\frac{{{\varepsilon }_{0}}AK}{4d}=\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{3}{4}+\frac{K}{4} \right]\] Or \[C'=\frac{C}{4}(K+3)\]        \[\left[ \therefore C=\frac{A{{\varepsilon }_{0}}}{d} \right]\]