A) \[\frac{3}{2}\frac{Qd}{{{\varepsilon }_{0}}A}\]
B) \[\frac{Qd}{{{\varepsilon }_{0}}A}\]
C) \[\frac{2}{3}\frac{Qd}{{{\varepsilon }_{0}}A}\]
D) \[\frac{3Qd}{{{\varepsilon }_{0}}A}\]
Correct Answer: C
Solution :
The arrangement of the plates can be rearranged as two capacitors connected in parallel. Equivalent capacitance of the system \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=\frac{{{\varepsilon }_{0}}A}{d}+\frac{{{\varepsilon }_{0}}A}{2d}=\frac{3{{\varepsilon }_{0}}A}{2d}\] Hence\[Q={{C}_{eq}}(V-0)=\left( \frac{3{{\varepsilon }_{0}}A}{2d} \right)V\Rightarrow V=\frac{2Qd}{3{{\varepsilon }_{0}}A}\]You need to login to perform this action.
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