Electrostatics > The potential of an electric field. Conductors in a constant electric field > Problem 6.3.12

Statement

$6.3.12.$ a. Two parallel oppositely charged metal plates are located at a distance of 1 cm from each other, much smaller than the size of the plates. The surface charge density of the plates is $\pm$3 CGS/cm$^2$. Determine the potential difference between the plates in CGS and SI.
b. Two parallel differently charged metal plates are located at a distance of 5 cm from each other, much smaller than the size of the plates. The surface charge density of the plates is $\pm$10$^{−10}$ C/cm$^2$. Determine the potential difference between the plates in CGS and SI.

Solution

Let´s suppose the upper plate with density $+\sigma$ and the lower one with density $-\sigma$, each plate generates an electric field of modular value $E = \frac{\sigma}{2\varepsilon_0}$. Then, the net field between them is $E_n = \frac{\sigma}{\varepsilon_0}$ because vectors are summed up (they have the same direction). Finally,
$V = -\int_{r}^{0} \vec{E} \cdot d\vec{r}$
$V(r) = \frac{\sigma}{\varepsilon_0}r$
a) Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$
$V = 4\pi \times 3\; \rm{\frac{esu}{cm^2}}\times 1\; \rm{cm}$
$V \simeq 37.7\; \rm{CGS}\; (\rm{statvolt})$
Taking in account that 1 statvolt = 299.792458 V,
$V \simeq 37.7\times 299.792458\; \rm{V} \simeq 11 300\; \rm{V} \simeq 11.3\; \rm{kV}$

b) For $r = 5$ cm, and 1 C = 2.9979$\times$10$^9$ esu,
$V =4\pi \times 10^{-10}\times 2,9979\times 10^9\; \rm{\frac{esu}{cm^2}}\times 5\; \rm{cm}$
$V = 18,85\; \rm{CGS}\; (\rm{statvolt})$
and in the SI,
$V = \frac{10^{-6}\; \rm{\frac{C}{m^2}}}{8.85\times10^{-12}\; \rm{\frac{C^2}{N\; m^2}}}\times 5\times10^{-2}\; \rm{m} \simeq 5650\; \rm{V} = 5.65\; \rm{kV}$

Answer

a) $V$ = 37.7 CGS, $V$ = 11.3 kV
b) $V$ = 18.85 CGS, $V$ = 5.65 kV