Question

An isolated $5.0\mu \mathrm{F}$ parallel-plate capacitor has $4.0\mathrm{mC}$ of charge. An external force changes the distance between the electrodes until the capacitance is $2.0\mu \mathrm{F}$. How much work is done by the external force?

Short answer

The amount of work is done by the external force is$2.4J$

Step-by-step solution

Given Information

Isolated parallel-plate Capacitor $=5.0\mu F$

Charge $=4.0mC$

Capacitance$=2.0\mu F$

Explanation

As long as the energy stored in the capacitor is different, a corresponding amount of work will be done. If we consider the latter as an equation, we see that this is necessary.

$E=\frac{q^2}{2C}$

Therefore, our work will be

$W=E_2-E_1=\frac{q^2}{2}\left(\frac{1}{C_2}-\frac{1}{C_1}\right)$

Which we can simplify as

$W=\frac{q^2\left(C_1-C_2\right)}{2C_1{C}_2}$

In our numerical case, we will have

$W=\left(\frac{{\displaystyle 0.004J^2\times 3·{10}^{-6}\mu F}}{{\displaystyle 2\times 5·{10}^{-6}\mu F\times 2·{10}^{-6}\mu F}}\right)$

Multiply the expression,

$=2.4\mathrm{J}$

Therefore, the amount of work is done by the external force is$2.4J$