JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance\[{{C}_{0}}\]. Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area \[\frac{1}{3}\operatorname{A},\] dielectric constant 2 K and another with area \[\frac{2}{3}\operatorname{A},\] and dielectric constant K. If the capacitance of this new capacitor is C then \[C/{{C}_{0}}\]is: [JEE ONLINE 25-04-2013]

A) \[1\]

B) \[\frac{4}{3}\]

C) \[\frac{2}{3}\]

D) \[\frac{1}{3}\]

Correct Answer: B

Solution :

[b] \[{{C}_{0}}=\frac{k{{\in }_{0}}A}{d}\]

\[C=\frac{k{{\in }_{0}}2}{3d}+\frac{2k{{\in }_{0}}A}{3d}=\frac{4}{3}\frac{k{{\in }_{0}}A}{d}\]

\[\therefore \]\[\frac{C}{{{C}_{0}}}=\frac{\frac{4}{3}\frac{k{{\in }_{0}}A}{d}}{\frac{k{{\in }_{0}}A}{d}}=\frac{4}{3}\]