Question

A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference of ∆V0. The charging battery is then disconnected (so the capacitor is electrically isolated) and the plates are pulled apart until their separation is 3d. Derive expressions in terms of A, d, and ∆V0 for (a) the new potential difference,(b) the initial and final stored energy, and (c) the work required to separate the plates.

Answer 1

(a) The initial potential difference between the plates:

$\varDelta V_0=\frac{q}{C},$

where q - the charge stored in the capacitor;

C - the capacitance of the capacitor.

Because the capacitor is electrically isolated, the charge stored in it remains constant. The new potential difference between the plates:

$\varDelta V_0'=\frac{q}{C'},$

where C' - the capacitance of the capacitor after increasing the plates separation.

The capacitance of the parallel plate capacitor is

$C=\varepsilon_r \varepsilon_0 \frac {A}{d},$

where $\varepsilon_r$ - the relative permittivity of the material between plates (for air $\varepsilon_r=1$ );

$\varepsilon_0$ - absolute dielectric permittivity of vacuum;

A - the area of the plate;

d - the plates separation.

If the new plates separation is  $d'=3d$, then $C'=\frac{C}{3}.$

So, the new potential difference:

$\varDelta V_0'=\frac{q}{C'}=\frac{3q}{C}=3\varDelta V_0.$

(b) The initial stored energy is

$W=\frac{C\varDelta V_0^2}{2}=\frac{\varepsilon_0 A\varDelta V_0^2}{2d}.$

The finak stored energy is

$W'=\frac{C'\varDelta V_0'^2}{2}=\frac{\varepsilon_0 A(3\varDelta V_0)^2}{2d'}=\frac{3\varepsilon_0 A\varDelta V_0^2}{2d}.$

(c) The work required to separate the plates equal to difference between the stored energies after and before increasing the plates separation:

$A=W'-W=\frac{3\varepsilon_0 A\varDelta V_0^2}{2d}-\frac{\varepsilon_0 A\varDelta V_0^2}{2d}=\frac{2\varepsilon_0 A\varDelta V_0^2}{2d}.$

Answer: (a) $3\varDelta V_0.$ (b) $\frac{\varepsilon_0 A\varDelta V_0^2}{2d}$ and $\frac{3\varepsilon_0 A\varDelta V_0^2}{2d}$ (c) $\frac{2\varepsilon_0 A\varDelta V_0^2}{2d}$