The Electric Field of a Charged Spherical Shell

The Electric Field of a Charged Spherical Shell

A spherical conducting shell of inner radius a and outer radius b carries a total charge +Q distributed on the surface of a conducting shell (Fig. 15.29a). The quantity Q is taken to be positive.

(a) Find the electric field in the interior of the conducting shell, for r < a, and

(b) the electric field outside the shell, for r > b.

(c) If an additional charge of -Q is placed at the center, find the electric field for r > b.

The Correct Answer and Explanation is :

Let’s break down the problem step by step and solve for the electric field at different points:

Part (a) – Electric Field Inside the Conducting Shell (for r<ar < a)

The spherical conducting shell carries a total charge +Q+Q distributed on its outer surface. A key property of conductors is that the electric field inside a conductor (and in the hollow region) is zero.

This is because charges within a conductor rearrange themselves in response to electric fields, such that the electric field inside the conducting material (the conductor itself) cancels out. In the case of a spherical shell, the charges reside only on the outer surface of the conductor. Hence, within the shell, the electric field is zero.

Therefore, the electric field for r<ar < a is: E=0forr<aE = 0 \quad \text{for} \quad r < a

Part (b) – Electric Field Outside the Shell (for r>br > b)

Now, for points outside the spherical shell, we can treat the shell as if all the charge +Q+Q were concentrated at the center of the shell, according to Gauss’s Law.

Consider a spherical Gaussian surface with radius rr such that r>br > b. Since the charge is distributed uniformly on the surface of the conducting shell, the electric field outside the shell is equivalent to the field produced by a point charge +Q+Q located at the center.

Using Gauss’s Law, which states: ∮E⋅dA=Qencϵ0\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}

where QencQ_{\text{enc}} is the charge enclosed by the Gaussian surface, we get: E(4πr2)=Qϵ0E(4 \pi r^2) = \frac{Q}{\epsilon_0}

Solving for the electric field EE: E=Q4πϵ0r2forr>bE = \frac{Q}{4 \pi \epsilon_0 r^2} \quad \text{for} \quad r > b

Part (c) – Electric Field Outside the Shell with an Additional Charge −Q-Q at the Center (for r>br > b)

Now, an additional charge of −Q-Q is placed at the center of the spherical shell. This charge will produce an electric field that must be considered along with the charge on the shell.

The total charge enclosed by a Gaussian surface outside the shell, for r>br > b, will now be the sum of the charge on the shell +Q+Q and the charge at the center −Q-Q, which totals to zero. Thus, the total charge enclosed is zero.

Using Gauss’s Law again for the spherical Gaussian surface: E(4πr2)=Qencϵ0=0E(4 \pi r^2) = \frac{Q_{\text{enc}}}{\epsilon_0} = 0

Hence, the electric field outside the shell with the central charge −Q-Q becomes: E=0forr>bE = 0 \quad \text{for} \quad r > b

Summary of Electric Fields:

• For r<ar < a (inside the shell): E=0E = 0
• For r>br > b (outside the shell, without the central charge): E=Q4πϵ0r2E = \frac{Q}{4 \pi \epsilon_0 r^2}
• For r>br > b (outside the shell, with the central charge −Q-Q): E=0E = 0

Explanation:

1. Conducting Shell: The spherical conducting shell ensures that the electric field inside the shell is zero. Conductors shield their interior from electric fields by rearranging the charges on their surface.
2. Gauss’s Law: The electric field outside a spherically symmetric charge distribution behaves as if the entire charge is concentrated at the center, which is why the field for r>br > b is the same as that of a point charge.
3. Zero Field Outside with Central Charge: When an additional −Q-Q charge is placed at the center of the shell, the net charge enclosed by any Gaussian surface outside the shell is zero, leading to no electric field outside the shell. This demonstrates how the configuration cancels out the field despite the presence of charges.