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.

Answer 2

This can also be reasoned out using the property that, seen from outside, a spherically symmetric charge distribution behaves exactly as if all its charge were concentrated at the centre, no matter how that charge is actually arranged inside.

1. Field outside is zero, net charge zero: Since the central charge Q must ultimately show up somewhere on the shell for the conductor's material to have zero internal field, the outer surface cannot end up neutral, and since there is nonzero enclosed charge overall, the field outside cannot be zero either.
2. Field is \( \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \), net charge \( +Q \): The central charge \( +Q \) induces \( -Q \) on the inner surface of the shell to cancel the field within the conducting material, and since the shell started with zero net charge, this forces \( +Q \) to appear on the outer surface. From outside, this outer \( +Q \) is spread symmetrically over a sphere, so it behaves exactly like a point charge \( +Q \) sitting at the centre, giving a field of \( \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \) at distance \( r \). This matches both parts of the option.
3. Field is \( \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \), net charge zero: The field expression here is correct, but if the outer surface truly carried zero net charge, there would be nothing left to produce any field outside at all, which contradicts the very field value stated in this option; the two parts are inconsistent with each other.
4. Field is zero, net charge \( +Q \): Having \( +Q \) sitting on a spherical outer surface would necessarily produce a field outside proportional to \( \frac{Q}{r^2} \), by the same symmetry that governs a solid sphere's external field; a nonzero net charge cannot give zero field outside, so these two parts also contradict each other.

Only the option combining an outer surface charge of \( +Q \) with an external field of \( \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \) is internally consistent with how conductors shield and redistribute charge.

Therefore, the correct answer is E = (1/4\(\pi\epsilon_0\))(Q/r\(^2\)), Net charge = +Q.