Answer to Question #129104 in Physics for Qwerty

Question

A sphere is charged to 20 µC. Find the potential at the following points. A, B and C (A point is at the center of the sphere, B is at the surface and C is 4cm from the surface.)

Answer 1

1) Let's choose the spherical shell inside the charged shere (drawn with thin line). Then, according to the Gauss's theorem, the flux of electrical field through this shell will be:

$$\Phi =ES = 4\pi k q_{inside}$$

where $S$ is the area of the shell surface, $k = 9\times 10^9 N\cdot m^2/C^2$ is the constant, and $q_{inside}$ is the charge inside the shell. Since the sphere is empty ( $q_{inside} = 0$ ), then $E = 0$ in any point inside the shere. Particulary, $E_A = 0.$

As far, as the potential is connected with the electric field in the following way:

$$E = -\dfrac{d\varphi}{dr}$$

the potintial in the point $A$ will be constant $\varphi_A = const$.

2) Let's choose the spherical shell outside the charged shere (drawn with thin line) that passes through the point $C$.

Then, according to the Gauss's theorem, the flux of electrical field through this shell will be:

$$\Phi =E_CS = 4\pi k q$$

where $S = 4\pi r^2$ is the area of the shell surface and $q = 20\times 10^{-6}C$ is the charge enclosed by this surface (the charge of the thick sphere). Hence, obtain:

$$E_C\cdot 4\pi r^2 = 4\pi kq\\ E_C = k\dfrac{q}{r^2}$$

The potential will be:

$$\varphi_C = -\int E_C dr = -\int k\dfrac{q}{r^2}dr = k\dfrac{q}{r} + C$$

we can choose the contstan $C$ to be zero if we set the potential be zero at large distance from the sphere. Thus:

$$\varphi_C = k\dfrac{q}{r} = k\dfrac{q}{R+4cm}$$

The distance $r$ can be considered as $r = R+4cm$, where $R$ is the radius of the charged sphere.

3) Then, the potential on the sphere surface is:

$$\varphi_B = k\dfrac{q}{R}$$

where $R$ is the radius of sphere.

Since the potential should be continuous, the potential inside the sphere should be equal to the potential on the surface. Thus:

$$\varphi_A = const = \varphi_B = k\dfrac{q}{R}$$

4) Let's summarize the results:

$$\varphi_A = k\dfrac{q}{R}\\ \space\\ \varphi_B=k\dfrac{q}{R}\\ \space\\ \varphi_C=k\dfrac{q}{R+4cm}\\$$

As far as the radius $R$ of the charged sphere is unknown, we can not give the final numerical answer. But let's come as close as we can:

$$\varphi_A = 9\times 10^9\cdot \dfrac{20\times 10^{-6}}{R} = 1.8\times10^5 \cdot \dfrac{1}{R}\space [V]\\ \space\\ \varphi_B=1.8\times10^5 \cdot \dfrac{1}{R}\space [V]\\ \space\\ \varphi_C=1.8\times10^5 \cdot \dfrac{1}{R+0.04m}\space [V]\\$$

Substituting the radius $R$ of the charged sphere in meters, obtain the final answer.

Answer.

$\varphi_A = 1.8\times10^5 \cdot \dfrac{1}{R}\space [V]\\ \space\\ \varphi_B=1.8\times10^5 \cdot \dfrac{1}{R}\space [V]\\ \space\\ \varphi_C=1.8\times10^5 \cdot \dfrac{1}{R+0.04m}\space [V]\\$