Question

(II) It takes 18 J of energy to move a 0.30-mC charge from one plate of a $\mathbf{15}\mu \mathbf{F}$ capacitor to the other. How much charge is on each plate?

Short answer

The charge on each plate is$0.90\mathrm{C}$.

Step-by-step solution

Understanding work done to move the charge

The potential difference may be defined as the amount of work done in moving a point charge of 1 C from one place to another.

It can be written as,

$V=\frac{W}{Q}$ … (i)

Here, V is the potential difference, W is the work done and Q is the charge.

Given Data

The work done is,$W=18\mathrm{J}$.

The amount of charge is,$q=0.30\mathrm{m}\mathrm{C}$.

The capacitance is,$C=15\mu \mathrm{F}$.

Determination of the charge on each plate

From equation (i), the work done is given as,

$\begin{array}{rl}W& =qV\\ W& =q\left(\frac{Q}{C}\right)\\ Q& =\frac{C\times W}{q}\end{array}$

Substitute the values in the above expression.

$\begin{array}{rl}Q& =\frac{\left(15\mu \mathrm{F}\times \frac{{10}^{-6}\mathrm{F}}{1\mu \mathrm{F}}\right)\left(18\mathrm{J}\right)}{\left(0.3\mathrm{m}\mathrm{C}\times \frac{{10}^{-3}\mathrm{C}}{1\mathrm{m}\mathrm{C}}\right)}\\ Q& =0.90\mathrm{C}\end{array}$

Thus, the charge on each plate is $0.90\mathrm{C}$.