Question

A dielectric of thickness 5 cm and dielectric constant 10, is introduced in between the plates of a parallel plate capacitor having plate area 500 cm2 and separation between plates 10 cm. The capacitance of the, capacitor is $ (If\,{{\varepsilon }_{0}}=8.8\times {{10}^{-12}}\,SI\,units): $

• A. 8 pF
• B. 6 pF
• C. 4 pF
• D. 20 pF

Answer 1

Answer: (A)

The capacitance of parallel plate capacitor when dielectric constant introduced between the plates is $ C=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{K} \right)} $ $ \left( \begin{align} & \text{Here}:\,K=10,t=5cm,\,A=500\,c{{m}^{2}}, \\ & d=10cm,\,{{\varepsilon }_{0}}=8.8\times {{10}^{-12}}\text{SI unit} \\ \end{align} \right) $ Putting the given value in above equation $ C=\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{10\times {{10}^{-2}}-5\times {{10}^{-2}}\left( 1-\frac{1}{10} \right)} $ $ =\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{10\times {{10}^{-2}}-4.5\times {{10}^{-12}}} $ $ =\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{5.5\times {{10}^{-12}}} $ $ =8\times {{10}^{-12}}F=8pF $