Question

When a potential difference of 150 V is applied to the
plates of a parallel-plate capacitor, the plates carry a surface
charge density of 30.0 nC/cm2
. What is the spacing
between the plates?

Answer 1

When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates?

Solution.

The capacitance of the parallel-plate capacitor is equal to the following:

$$C = \frac {\varepsilon_ {0} A}{d},$$

where $C$ is the capacitance, $A$ is the area of overlap of the two plates, $\varepsilon_0$ is the electric constant ($\varepsilon_0 \approx 8.85 \cdot 10^{-12} \, \text{F/m}$), $d$ is the spacing between the plates.

On the other hand, the capacitance is given by:

$$C = \frac {q}{U},$$

where $q$ is the charge of one plate, $U$ is the potential difference between the plates. The charge may be expressed in terms of the surface charge density $\sigma$:

$$q = \sigma A.$$

Considering all the formulas above we have:

$$\frac {\varepsilon_ {0} A}{d} = \frac {\sigma A}{U}; d = \frac {\varepsilon_ {0} U}{\sigma} = \frac {8.85 \cdot 10^{-12} \, \frac {\text{F}}{\text{m}} \cdot 150 \, \text{V}}{30 \cdot 10^{-5} \, \frac {\text{C}}{\text{m}^2}} = 4.43 \, \mu\text{m}.$$

Answer: $4.43 \, \mu\text{m}$.