Pulling apart a capacitor

The capacitance of a capacitor is given by $C = \frac{k \epsilon_{0} A }{d}$ . From this formula we see that if the distance between the plates is doubled, then the capacitance becomes half. However, the charge on the capacitor remains the same since it is trapped on the metal plates and cannot escape.

There are three ways to write the stored energy of a capacitor:

$\frac{1}{2} \frac{Q^{2}}{C}, \quad \frac{1}{2} CV^{2}, \quad \frac{1}{2} QV$

We will proceed with the first way since we know the charge and the capacitance of capacitor and how they change as the distance between the plates changes.

The initial stored energy of the capacitor is $E_{i} = \frac{1}{2} \frac{Q^{2}}{C}$ . After pulling apart the plates, the capacitance becomes half and the charge remains the same. Therefore the final energy stored in the capacitor is $E_{f} = \frac{1}{2} \frac{Q^{2}}{C/2}$ .

The change in energy of the capacitor is $\Delta E = E_{f} - E_{i} =\frac{1}{2} \frac{Q^{2}}{C}$ . We must supply this amount of energy externally in order to successfully pull apart the plates.

We are given that the charge $Q = 50 \ \mu\text{C} = 50 \times 10^{-6} \text{ C}$ , and capacitance $C = 10 \mu \text{F} = 10 \times 10^{-6} \text{ F}$ . When we substitute these values in the equation, we get $\Delta E = \dfrac{1}{2} \times \dfrac{ 50 \times 10^{-6} \times 50 \times 10^{-6}}{ 10 \times 10^{-6}} = \boxed{125 \ \mu \text{J}}$ $_\square$