Two concentric rings placed in a gravity free region in yz-plane one of radius R carries `a + Q` and second of radius `4R` and charge `–8Q` distributed uniformly over it. Find the velocity with which a point charge of mass m and charge `+q` should be projected from a a distance `3R` from the centre of rings on its axis so that it will reach to the centre of the rings.

Explanation

To solve the problem, we will use the principle of conservation of energy. We need to find the velocity \( v \) with which a point charge \( +q \) should be projected from a distance \( 3R \) from the center of the concentric rings so that it reaches the center of the rings.

### Step-by-Step Solution:

1. **Identify the Charges and Distances:**
- The inner ring has a charge \( +Q \) and radius \( R \).
- The outer ring has a charge \( -8Q \) and radius \( 4R \).
- The point charge \( +q \) is projected from a distance \( 3R \) from the center of the rings.

2. **Calculate Initial Potential Energy (PE_initial):**
- The distance from the charge \( +q \) to the charge \( +Q \) when it is at \( 3R \):
\[ d_1 = \sqrt{(3R)^2 + R^2} = \sqrt{9R^2 + R^2} = \sqrt{10R^2} = \sqrt{10}R \]
- The potential energy due to the inner ring:
\[ PE_{1} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Qq}{\sqrt{10}R} \]

- The distance from the charge \( +q \) to the charge \( -8Q \):
\[ d_2 = \sqrt{(3R)^2 + (4R)^2} = \sqrt{9R^2 + 16R^2} = \sqrt{25R^2} = 5R \]
- The potential energy due to the outer ring:
\[ PE_{2} = \frac{1}{4\pi\epsilon_0} \cdot \frac{-8Qq}{5R} \]

- Total initial potential energy:
\[ PE_{initial} = PE_{1} + PE_{2} = \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{\sqrt{10}R} - \frac{8Qq}{5R} \right) \]

3. **Calculate Final Potential Energy (PE_final):**
- When the charge \( +q \) reaches the center of the rings, the distances are:
- To the inner ring: \( R \)
- To the outer ring: \( 4R \)

- The potential energy at the center:
\[ PE_{final} = \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{R} - \frac{8Qq}{4R} \right) = \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{R} - \frac{2Qq}{R} \right) = \frac{1}{4\pi\epsilon_0} \left( -\frac{Qq}{R} \right) \]

4. **Apply Conservation of Energy:**
- The total mechanical energy is conserved:
\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]
- The initial kinetic energy \( KE_{initial} = \frac{1}{2}mv^2 \) and \( KE_{final} = 0 \) (since it stops at the center):
\[ \frac{1}{2}mv^2 + PE_{initial} = 0 + PE_{final} \]

5. **Set Up the Equation:**
\[ \frac{1}{2}mv^2 = -PE_{initial} + PE_{final} \]
\[ \frac{1}{2}mv^2 = -\left( \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{\sqrt{10}R} - \frac{8Qq}{5R} \right) \right) + \left( \frac{1}{4\pi\epsilon_0} \left( -\frac{Qq}{R} \right) \right) \]

6. **Simplify and Solve for \( v \):**
- Rearranging and simplifying the equation will allow us to isolate \( v \):
\[ \frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{\sqrt{10}R} - \frac{8Qq}{5R} + \frac{Qq}{R} \right) \]
- After simplification, solve for \( v \):
\[ v = \sqrt{\frac{2}{m} \cdot \frac{1}{4\pi\epsilon_0} \left( \frac{Qq}{\sqrt{10}R} - \frac{8Qq}{5R} + \frac{Qq}{R} \right)} \]