The capacitance of capacitor will be:
A) \[\frac{{{\varepsilon }_{0}}A\,({{K}_{1}}+{{K}_{2}})}{2\,d}\]
B) \[\frac{{{\varepsilon }_{0}}A}{2\,d}\left( \frac{({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}} \right)\]
C) \[\frac{{{\varepsilon }_{0}}}{d}\left( \frac{({{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}})} \right)\]
D) \[\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{({{K}_{1}}+{{K}_{2}})}{{{K}_{1}}{{K}_{2}}} \right)\]
Correct Answer: A
Solution :
Key Idea: In each capacitor, the area of the plate will be\[\frac{A}{2}\]. In the given combination, the arrangement is equivalent to two capacitors connected in parallel. Also in each capacitor the area of the plate will be \[\frac{A}{2}\]. Therefore, equivalent capacitance \[C={{C}_{1}}+{{C}_{2}}\] \[=\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}A/2}{d}\] \[=\frac{{{\varepsilon }_{0}}A}{2\,d}\,({{K}_{1}}+{{K}_{2}})\]
You need to login to perform this action.
You will be redirected in
3 sec
