Question

Two parallel plate capacitors ($AB\&CD$) differ only in the spacing between their plates(very thin).
Capacitor $AB$ has a spacing of $5\text{mm}$ and a capacitance of $20\text{pF}$, the capacitor $CD$, has a spacing of $2\text{mm}$.
Plates $A$ and $C$ carry charges of $+1\text{nC}$, while $B$ and $D$ each carry $-1\text{nC}$.
If the capacitor $CD$ is slid centrally between and parallel to the plates of $AB$ without touching them (as shown below), what is the potential difference $V_{AB}$ ?

• A. $70\text{V}$
• B. $50\text{V}$
• C. $40\text{V}$
• D. None of these

Answer 1

The correct option is A $70\text{V}$

Charge re-distribution on the plates is as shown below:


$V_{AC}=\frac{Q}{C_{AC}}=Q\frac{d_{CA}}{{ϵ}_{0}A}$
$V_{DB}=\frac{Q}{C_{DB}}=Q\left[\frac{d_{DB}}{{ϵ}_{0}A}\right]$
$V_{AC}+V_{DB}=Q\frac{d_{CA}+d_{DB}}{{ϵ}_{0}A}$
$V_{AC}+V_{DB}=Q\frac{\left(3\right)}{{ϵ}_{0}A}=Q\frac{5}{{ϵ}_{0}A}\times \frac{3}{5}=\frac{1\text{nC}}{20\text{pF}}\times \frac{3}{5}$
$V_{AC}+V_{DB}=30\text{V}...\left(1\right)$

$V_{CD}=\frac{Q}{C_{CD}}=\frac{2\text{nC}}{20\text{pF}\times \frac{5}{2}}=40\text{V}...\left(2\right)$
On adding (1) and (2), we get $V_{AB}=30\text{V}+40\text{V}=70\text{V}$