If the distance between the plates of a parallel plate capacitor of capacity 10 μ F is doubled, then new capacity will be

Question

If the distance between the plates of a parallel plate capacitor of capacity 10 μ F is doubled, then new capacity will be (A)  5 μ F  (B)  20 μ F  (C)  10 μ F  (D)  15 μ F

Tardigrade Admin · Accepted Answer

In parallel plate capacitor C=ε0 (A/d) ...(i) Now d'=2 d Hence, C'=(ε0 A/d')=(ε0 A/2 d) ...(ii) From Eqs. (i) and (ii) (C'/C)=(1/2) C'=(C/2) Putting C =10 μ F C'=(10/2)=5 μ F

Q. If the distance between the plates of a parallel plate capacitor of capacity $10 μ F$ is doubled, then new capacity will be

$5 μ F$

$20 μ F$

$10 μ F$

$15 μ F$

In parallel plate capacitor
$C = ε_{0} \frac{A}{d} ... (i)$
Now $d^{'} = 2 d$
Hence, $C^{'} = \frac{ε _{0} A}{d ^{'}} = \frac{ε _{0} A}{2 d} ... (ii)$
From Eqs. (i) and (ii)
$\frac{C ^{'}}{C} = \frac{1}{2}$
$C^{'} = \frac{C}{2}$
Putting $C = 10 μ F$
$C^{'} = \frac{10}{2} = 5 μ F$

Q. If the distance between the plates of a parallel plate capacitor of capacity 10μF is doubled, then new capacity will be

5μF

20μF

10μF

15μF