Question

A solid conducting sphere of radius \(R\) is centered about the origin of an \(x y z\) -coordinate system. A total charge \(Q\) is distributed uniformly on the surface of the sphere. Assuming, as usual, that the electric potential is zero at an infinite distance, what is the electric potential at the center of the conducting sphere? a) zero b) \(Q / \epsilon_{0} R\) c) \(Q / 2 \pi \epsilon_{0} R\) d) \(Q / 4 \pi \epsilon_{0} R\)

Short answer

Answer: (d) \(Q / 4 \pi \epsilon_{0} R\)

Step-by-step solution

Apply Gauss's law

Since the sphere is conducting, the electric field inside the sphere is zero. We can now apply Gauss's law to find the electric field outside the sphere. Gauss's law states that: \(\oint \vec{E} \cdot d \vec{A} = \frac{Q_{inside}}{\epsilon_0}\), where \(Q_{inside}\) is the total charge enclosed within the Gaussian surface. In this case, the Gaussian surface should be a sphere centered at the origin with a radius \(r > R\). Since the electric field is radially outward, it is parallel to the infinitesimal area vector \(d \vec{A}\), so \(\vec{E} \cdot d \vec{A} = E dA\). Therefore, Gauss's law becomes: \(\oint EdA = \frac{Q}{\epsilon_0}\).

Find the electric field outside the sphere

As the electric field is constant on the Gaussian surface because of symmetry, the integral of \(EdA\) is equal to \(E\) times the surface area of the Gaussian sphere, which can be calculated as \(4 \pi r^2\). So we have \(\oint EdA = E \times 4 \pi r^2 = \frac{Q}{\epsilon_0}\). Solving for \(E\), we get: \(E = \frac{Q}{4 \pi \epsilon_0 r^2}\).

Find the electric potential

Now that we know the electric field outside the sphere, we can find the electric potential at a distance \(r\) from the center. The relation between the electric field and electric potential \(\phi\) is: \(d \phi = - E dr\). To find the electric potential at the center of the sphere (\(r = 0\)), we need to integrate this expression from the surface of the sphere to the center, with the boundary condition of electric potential being zero at infinity. \(\phi(0) - \phi(R) = - \int_{R}^{0} \frac{Q}{4 \pi \epsilon_0 r^2} dr\)

Evaluate the integral

Evaluating this integral, we get: \(\phi(0) - \phi(R) = \frac{Q}{4 \pi \epsilon_0 } \left( \frac{1}{R} - \frac{1}{0} \right)\) Since the potential at infinity is zero, we get: \(\phi(0) = -\phi(R) = \frac{Q}{4 \pi \epsilon_0 R}\) Thus, the electric potential at the center of the sphere is option (d): \(Q / 4 \pi \epsilon_{0} R\).