Question

A parallel plate capacitor of value 1.77 $ \mu $ F is to be designed using a dielectric material (dielectric constant = 200), breakdown strength of $ 3\times {{10}^{6}} $ V/m. In order to make such a capacitor which can withstand a potential difference of 20 V across the plates, the separation between the plates d and area A of the plates respectively are

• A. $ 6.6\times {{10}^{-6}}m,{{10}^{3}}{{m}^{2}} $
• B. $ 6.6\times {{10}^{-5}}m,{{10}^{4}}{{m}^{2}} $
• C. $ 6.6\times {{10}^{-4}}m,\text{ }{{10}^{5}}{{m}^{2}} $
• D. $ 6.6\times {{10}^{-6}}m,\text{ }{{10}^{2}}{{m}^{2}} $

Answer 1

Answer: (A)

We knows, electric field is given by
$ E=\frac{V}{d} $
$ {{E}_{\max }}=\frac{V}{{{d}_{\min }}} $
$ {{d}_{\min }}=\frac{V}{{{E}_{\max }}} $
Substitute $ V=20\text{ }V $ and $ {{E}_{\max }}=3\times {{10}^{6}}V $
$ {{d}_{\min }}=\frac{20}{3\times {{10}^{6}}}=6.6\times {{10}^{-6}}m $
We knows, the capacitance is given by
$ C=\frac{k{{\varepsilon }_{0}}A}{d} $
$ \frac{A}{d}=\frac{C}{k{{\varepsilon }_{0}}} $
$ A=\frac{C}{k{{\varepsilon }_{0}}} $
Putting $ C=1.77\text{ }\mu F=1.77\times {{10}^{-6}}C, $
$ d=6.6\times {{10}^{-6}},k=200, $
$ {{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{2}} $
$ A=\frac{1.77\times {{10}^{-6}}\times 6.6\times {{10}^{-6}}}{200\times 8.85\times {{10}^{-12}}}={{10}^{3}}{{m}^{2}} $