Question

An isolated charged conducting sphere of radius 12.0 cm
creates an electric field of 4.90 * 104
N/C at a distance
3.
21.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance?

Answer 1

Question 26214

1) First, let us use divergence theorem (Gauss-Ostrogradsky theorem) to find electric field outside the sphere as a function of distance from center of sphere (Let $r$ be that distance, and $\mathbf{R}$ be the radius of sphere):

$$\oint \vec{E} \, \mathrm{d}S = \frac{Q}{\varepsilon_0}$$ $$4\pi E r^2 = 4\pi \sigma \frac{R^2}{\varepsilon_0} \Rightarrow E(r) = \frac{r^2 \sigma}{R^2 \varepsilon_0}.$$

From here obtain surface charge density ($r = 12 \, \text{cm}$, $R = 321 \, \text{cm}$): $\sigma = \frac{E r^2 \varepsilon_0}{R^2} = 0.00031 \, \frac{C}{m^2}$.

2) Let us find first formula for spherical capacitor with spheres of radius $R$ and $R_1$ (R is fixed – radius of given sphere, $R_1$ is changeable):

By definition, $C = \frac{Q}{U}$, and $U = \oint \vec{E} \, \mathrm{d}S = \frac{Q}{4\pi \varepsilon_0} \int_{R}^{R_1} \frac{dr}{r^2} = \frac{Q}{4\pi \varepsilon_0} \left( \frac{1}{R} - \frac{1}{R_1} \right)$, so $C = \frac{4\pi \varepsilon_0}{\left( \frac{1}{R} - \frac{1}{R_1} \right)}$.

Finally, taking $R_1 \to \infty$, obtain capacitance of sphere: $C = 4\pi \varepsilon_0 R = 1.33 \cdot 10^{-11} \, \text{F}$.