Question

$n$ capacitors each of capacitance 2 $\mu F$ are connected in parallel and a potential difference of $200V$ is applied to the combination. The total charge on all the positive plates is $1$ Coulomb then $n$ is equal to :

• A. $3333$
• B. $3000$
• C. $2500$
• D. $25$

Answer 1

The correct option is C $2500$

In parallel connection potential across capacitor is same as in one capacitor = $Q=C\times V=2\times {10}^{-6}\times 200$
$=2\times {10}^{-6}\times 200$
$=4\times {10}^{-4}C$
total charge $=n\times 4\times {10}^{-4}$
$n\times 4\times {10}^{-4}=1$
$n=\frac{10000}{4}$
$n=2500$