A) 8 pF
B) 6 pF
C) 4 pF
D) 20 pF
Correct Answer: A
Solution :
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 \[C=\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{10\times {{10}^{-2}}-5\times {{10}^{-2}}\left( 1-\frac{1}{10} \right)}\] \[=\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{10\times {{10}^{-2}}-4.5\times {{10}^{-12}}}\] \[=\frac{8.8\times {{10}^{-12}}\times 500\times {{10}^{-4}}}{5.5\times {{10}^{-12}}}\] \[=8\times {{10}^{-12}}F=8pF\]
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