$6.3.12.$ 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, $\frac{1}{\varepsilon_0} = 4\pi$\
$6.3.12.$ 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.\
$6.3.12.$ 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.
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
### 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,\
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,\
$V = -\int_{r}^{0}\vec{E}\cdot d\vec{r}$\
$V = -\int_{r}^{0}\vec{E}\cdot d\vec{r}$\
$V(r) = \frac{\sigma}{\varepsilon_0}r$\
$V(r) = \frac{\sigma}{\varepsilon_0}r$\
a) Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\
a) Calculating, for $r$ = 1 cm, in CGS system. In this case, $\frac{1}{\varepsilon_0} = 4\pi$\
