A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 µC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

• The capacitance of a parallel plate capacitor is, $C=920pF$.
• The charge is,$Q=3.9\mu C$.
• The plate separation distance is, $d$.

Now, we know that the difference between the two plates of a capacitor depends on the change on the conducting plates and is given by:

${\mathrm{V}}_{\mathrm{ab}}=\frac{\mathrm{Q}}{\mathrm{C}}$

Substitute the$3.90\mu C$ for Q and 920 pF for C in the above equation.

$\begin{array}{rcl}{V}_{ab}& =& \frac{3.90\times {10}^{-6}C}{920\times {10}^{-12}F}\\ & =& 4240V\end{array}$

Therefore, the potential difference is 4240V.

We know that the potential difference is inversely proportional to the capacitance C, therefore

$V_{ab}\propto \frac{1}{C}$

Again, the capacitance is related to the separated distance by the relation

$C={\epsilon }_{\circ }\frac{A}{d}$

As shown the distance d is inversely proportional to capacitance:

$d\propto \frac{1}{C}$

From the above two equalities we get:

$d\propto V_{ab}$

Hence as the distance doubled the potential difference doubles and becomes:

$V_{ab}=8480\text{V}$

Therefore, the potential difference after the distance is doubles is 8480 V.

To calculate the work done required to double the distance d. We will use the equation:

$\begin{array}{rcl}W& =& U\\ & =& \frac{Q^2}{2C}\\ & & \end{array}$

Substitute the$3.9\mu C$ for Q and 920 pF for C in the above equation.

$\begin{array}{rcl}\frac{Q^2}{2C}& =& \frac{{(3.90\times {10}^{-6}\text{C})}^2}{2(920\times {10}^{-12}\text{F})}\\ & =& 0.00826\text{J}\end{array}$

Therefore, the work done is 0.00826 J.