Question

A charge of $+2·0\times {10}^{-8}\mathrm{C}$ is placed on the positive plate and a charge of $-1·0\times {10}^{-8}\mathrm{C}$ on the negative plate of a parallel-plate capacitor of capacitance $1·2\times {10}^{-3}\mu \mathrm{F}$. Calculate the potential difference developed between the plates.

Answer 1

The charge on the positive plate is q₁ and that on the negative plate is q₂.
Given:
$$\begin{gathered} q_1=2.0\times {10}^{-8}\mathrm{C} \\ {q}_2=-1.0\times {10}^{-8}\mathrm{C} \end{gathered}$$

Now,
Net charge on the capacitor = $\frac{\left(q_1-q_2\right)}{2}=1.5\times {10}^{-8}\mathrm{C}$

The potential difference developed between the plates is given by
$$\begin{gathered} q=VC \\ \Rightarrow V=\frac{1.5\times {10}^{-8}}{1.2\times {10}^{-9}}=12.5\mathrm{V} \end{gathered}$$