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