Question

A charge of $1$$\mu$C is given to one plate of a capacitors and a charge of $2$$\mu$C is given to the other plate of a $0.1$$\mu$F capacitor. Find the potential difference across the two plates of the capacitor:

• A. $5V$
• B. $10V$
• C. $15V$
• D. $30V$

Answer 1

Answer: (A)

$5V$

Charge distribution on each side of the plates as shown in figure below and here $Q_1>Q_2$.

The electric field at point A inside the plate is zero, So $\frac{1}{A{ϵ}_0}[Q_1-q-q-Q_2+q^{\prime }-q^{\prime }]=0\Rightarrow q=\frac{Q_1-Q_2}{2}$.

similarly for point B, $\frac{1}{A{ϵ}_0}[Q_1-q+q+Q_2-q^{\prime }-q^{\prime }]=0\Rightarrow q^{\prime }=\frac{Q_1+Q_2}{2}$.

Thus the charge on inner sides of plates is , $Q_{net}=q=\frac{Q_1-Q_2}{2}=\frac{2-1}{2}=0.5\mu C$.

Potential $V=\frac{Q_{net}}{C}=\frac{0.5}{0.1}=5V$