A) 5 \[\mu F\]
B) 20\[\mu F\]
C) 10\[\mu F\]
D) 15\[\mu F\]
Correct Answer: A
Solution :
The capacity of a parallel plate capacitor of plate area A, and distance d between the plates is given by \[C=\frac{{{\varepsilon }_{0}}A}{d}\] Given, \[{{C}_{1}}=10\mu F,{{d}_{1}}=d,{{d}_{2}}=2d\] \[\therefore \] \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{d}_{2}}}{{{d}_{1}}}\] \[\Rightarrow \] \[{{C}_{2}}=\frac{{{C}_{1}}{{d}_{1}}}{{{d}_{2}}}\] \[=\frac{10\times d}{2d}=5\mu F\] Hence, capacity decreases when plate distance increases.
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