Question

The two thin coaxial rings, each of radius 'a' and having charges +Q and -Q respectively are separated by a distance of 's'. The potential difference between the centres of the two rings is:

• A. \( \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \)
• B. \( \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{a} + \frac{1}{\sqrt{s^2 + a^2}} \right] \)
• C. \( \frac{Q}{2\pi\epsilon_0} \left[ \frac{1}{a} + \frac{1}{\sqrt{s^2 + a^2}} \right] \)
• D. \( \frac{Q}{2\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \)

Hint:

When calculating potential due to a charge distribution at a point, you need the distance from every part of the charge to that point. For a ring, all points on the circumference are equidistant from any point on its axis. Remember to use the superposition principle and sum the potentials algebraically.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the electric potential difference between the centers of two coaxial rings with equal and opposite charges.
Step 2: Key Formula or Approach:
1. The electric potential `V` at a point on the axis of a ring of charge `Q` and radius `a`, at a distance `x` from its center, is given by \( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{\sqrt{x^2 + a^2}} \).
2. The total potential at a point due to multiple charges is the algebraic sum of the potentials due to individual charges (Principle of Superposition).
3. We need to calculate the potential at the center of each ring and then find their difference.
Step 3: Detailed Explanation:
Let's place the ring with charge +Q at the origin (x=0) and the ring with charge -Q at x=s. Let the centers be C1 (at x=0) and C2 (at x=s).
Potential at the center of the first ring (V1 at C1):
- Potential at C1 due to the first ring (+Q) itself. Here, the distance `x` is 0. \[ V_{1, on-1} = \frac{1}{4\pi\epsilon_0} \frac{+Q}{\sqrt{0^2 + a^2}} = \frac{Q}{4\pi\epsilon_0 a} \] - Potential at C1 due to the second ring (-Q). The distance between the center C1 and any point on the second ring is \( \sqrt{s^2 + a^2} \). So, the potential at C1 due to the second ring is: \[ V_{1, on-2} = \frac{1}{4\pi\epsilon_0} \frac{-Q}{\sqrt{s^2 + a^2}} = -\frac{Q}{4\pi\epsilon_0 \sqrt{s^2 + a^2}} \] - Total potential at C1 is the sum: \[ V_1 = V_{1, on-1} + V_{1, on-2} = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \] Potential at the center of the second ring (V2 at C2):
- Potential at C2 due to the first ring (+Q). The distance is `s`. The distance from any point on the first ring to C2 is \( \sqrt{s^2 + a^2} \). \[ V_{2, on-1} = \frac{1}{4\pi\epsilon_0} \frac{+Q}{\sqrt{s^2 + a^2}} \] - Potential at C2 due to the second ring (-Q) itself (distance `x=0` from its own center). \[ V_{2, on-2} = \frac{1}{4\pi\epsilon_0} \frac{-Q}{\sqrt{0^2 + a^2}} = -\frac{Q}{4\pi\epsilon_0 a} \] - Total potential at C2 is the sum: \[ V_2 = V_{2, on-1} + V_{2, on-2} = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{\sqrt{s^2 + a^2}} - \frac{1}{a} \right] \] Potential Difference (\(\Delta V\)):
\[ \Delta V = V_1 - V_2 \] \[ \Delta V = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] - \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{\sqrt{s^2 + a^2}} - \frac{1}{a} \right] \] \[ \Delta V = \frac{Q}{4\pi\epsilon_0} \left[ \left(\frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}}\right) - \left(\frac{1}{\sqrt{s^2 + a^2}} - \frac{1}{a}\right) \right] \] \[ \Delta V = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} - \frac{1}{\sqrt{s^2 + a^2}} + \frac{1}{a} \right] \] \[ \Delta V = \frac{Q}{4\pi\epsilon_0} \left[ \frac{2}{a} - \frac{2}{\sqrt{s^2 + a^2}} \right] = \frac{2Q}{4\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \] \[ \Delta V = \frac{Q}{2\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \] Step 4: Final Answer:
The potential difference between the centers is \( \frac{Q}{2\pi\epsilon_0} \left[ \frac{1}{a} - \frac{1}{\sqrt{s^2 + a^2}} \right] \). This corresponds to option (D).