Field of a Charged Conducting Sphere

We place a total positive charge on a solid conducting sphere with radius (Fig. 1). Find at any point inside or outside the sphere.

As we discussed earlier in this section, all of the charge must be on the surface of the sphere. The charge is free to move on the conductor, and there is no preferred position on the surface; the charge is therefore distributed uniformly over the surface, and the system is spherically symmetric. To exploit this symmetry, we take as our Gaussian surface a sphere of radius centered on the conductor. We can calculate the field inside or outside the conductor by taking r  < R or r > R, respectively. In either case, the point at which we want to calculate E lies on the Gaussian surface.

Fig.1: Calculating the electric field of a conducting sphere with positive charge q. Outside the sphere, the field is the same as if all of the charge were concentrated at the center of the sphere.

The spherical symmetry means that the direction of the electric field must be radial; that’s because there is no preferred direction parallel to the surface, so E can have no component parallel to the surface. There is also no preferred orientation of the sphere, so the field magnitude can depend only on the distance from the center and must have the same value at all points on the Gaussian surface.

For r > R the entire conductor is within the Gaussian surface, so the enclosed charge is q. The area of the Gaussian surface is $4\pi r^2$, and E is uniform over the surface and perpendicular to it at each point. The flux integral $\oint E_{\perp}dA$ is then just $E.(4\pi r^2)$ and Eq. $\Phi_E=\oint \mathbf{E}.d\mathbf{A}=\frac{q_{encl}}{\epsilon_0}$ gives

$E.(4\pi r^2)=\frac{q}{\epsilon_0}$ and

$E=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$  (outside a charged conducting sphere)

This expression is the same as that for a point charge; outside the charged sphere, its field is the same as though the entire charge were concentrated at its center. Just outside the surface of the sphere, where r = R, 

$E=\frac{1}{4\pi \epsilon_0}\frac{q}{R^2}$ (at the surface of a charged conducting sphere)

CAUTION: Flux can be positive or negative Remember that we have chosen the charge to be positive. If the charge is negative, the electric field is radially inward instead of radially outward, and the electric flux through the Gaussian surface is negative. The electric-field magnitudes outside and at the surface of the sphere are given by the same expressions as above, except that denotes the magnitude (absolute value) of the charge.

For r < R we again have $E.(4\pi r^2)=\frac{Q_{encl}}{\epsilon_0}$. But now our Gaussian surface (which lies entirely within the conductor) encloses no charge, so $Q_{encl}$. The electric field inside the conductor is therefore zero.

We already knew that E = 0 inside a solid conductor (whether spherical or not) when the charges are at rest. Figure 1 shows E as a function of the distance r from the center of the sphere. Note that in the limit as R → 0, the sphere becomes a point charge; there is then only an “outside,” and the field is everywhere given by $E=q/(4\pi \epsilon_0r^2)$. Thus we have deduced Coulomb’s law from Gauss’s law.

We can also use this method for a conducting spherical shell (a spherical conductor with a concentric spherical hole inside) if there is no charge inside the hole. We use a spherical Gaussian surface with radius less than the radius of the hole. If there were a field inside the hole, it would have to be radial and spherically symmetric as before, so $E=Q_{encl}/(4\pi \epsilon_0r^2)$. But now there is no enclosed charge, so $Q_{encl}=0$ and E = 0  inside the hole.

Can you use this same technique to find the electric field in the region between a charged sphere and a concentric hollow conducting sphere that surrounds it?

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