Question

A charge Q is placed at the center of an uncharged, hollow, conducting spherical shell. What is the electric field at a point outside the shell at a distance r from the center, and what is the net charge on the outer surface of the shell?

• A. E = 0, Net charge = 0
• B. E = (1/4\(\pi\epsilon_0\))(Q/r\(^2\)), Net charge = +Q
• C. E = (1/4\(\pi\epsilon_0\))(Q/r\(^2\)), Net charge = 0
• D. E = 0, Net charge = +Q

Hint:

Remember the key properties of conductors in electrostatic equilibrium:
- Electric field inside a conductor is zero.
- Any net charge on a conductor resides entirely on its outer surface.
- Induced charges distribute to cancel internal fields or maintain overall charge neutrality.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for two things: the electric field outside an uncharged hollow conducting spherical shell with a charge Q at its center, and the net charge on the shell's outer surface.

Step 2: Key Formula or Approach:
1. Electric field inside a conductor: In electrostatic equilibrium, the electric field inside a conductor is zero.
2. Gauss's Law: \( \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0} \).
3. Conservation of Charge: For an isolated system, the total charge remains constant.

Step 3: Detailed Explanation:
Consider an uncharged, hollow, conducting spherical shell with a charge Q placed at its center.
1. Charge Distribution on the Shell:
* To maintain an electric field of zero inside the conducting material of the shell, a charge of -Q will be induced on the inner surface of the shell. This -Q charge perfectly cancels the electric field from the central +Q charge within the conductor's material.
* Since the conducting shell was initially uncharged, to maintain overall charge neutrality for the shell, a charge of +Q must appear on its outer surface.
* Thus, the net charge on the outer surface of the shell is +Q.
2. Electric Field at a point outside the shell (at distance r from the center):
* To find the electric field at a point outside the shell, we can draw a spherical Gaussian surface of radius \(r\) (where \(r\) is greater than the shell's outer radius) concentric with the shell.
* The total charge enclosed by this Gaussian surface is the sum of the central charge (+Q) and the charge on the outer surface of the shell (+Q). So, $Q_{enclosed} = +Q$. (The charge -Q on the inner surface is enclosed, but the +Q on the outer surface is also enclosed; effectively, the net charge is +Q from the center).
* Alternatively, the effect of an induced charge on the outer surface of a conductor is equivalent to placing that charge at the center for points outside. So, the electric field outside is due to the total net charge inside/on the shell system, which is +Q.
* Applying Gauss's Law: \( E (4\pi r^2) = \frac{Q}{\epsilon_0} \)
* So, the electric field $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.

Step 4: Final Answer:
The electric field at a point outside the shell is $(1/4\pi\epsilon_0)(Q/r^2)$, and the net charge on the outer surface of the shell is +Q.