Three large conducting plates carrying charges `Q,2Q and 3Q` on them, are placed parallel to each other as shown in the figure. If `U_(1) and U_(2)` denote electrostatic potential energy in the space between `1^(st)` and `2^(nd)` plane and `2^(nd)` and `3^(rd)` plate respectively as indicated, then `(U_(2))/(U_(1))` equals

Options

A. 1
B. 0
C. `2//3`
D. `3//2`

Answer

To solve the problem, we need to find the ratio of the electrostatic potential energy \( U_2 \) between the second and third plates to the electrostatic potential energy \( U_1 \) between the first and second plates.

### Step-by-Step Solution:

1. **Identify the Charges on the Plates**:
- Let the charges on the plates be \( Q_1 = Q \), \( Q_2 = 2Q \), and \( Q_3 = 3Q \).

2. **Determine the Electric Field Between the Plates**:
- The electric field \( E \) between two charged plates can be calculated using the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.
- For the first and second plates, the net electric field \( E_{1-2} \) can be calculated as:
\[ E_{1-2} = \frac{Q}{A \epsilon_0} + \frac{2Q}{A \epsilon_0} = \frac{3Q}{A \epsilon_0} \]
- For the second and third plates, the net electric field \( E_{2-3} \) is:
\[ E_{2-3} = \frac{2Q}{A \epsilon_0} + \frac{3Q}{A \epsilon_0} = \frac{5Q}{A \epsilon_0} \]

3. **Calculate the Potential Energy**:
- The electrostatic potential energy \( U \) in a region between two plates can be expressed as:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the potential difference.
- The potential difference \( V \) between two plates is given by:
\[ V = E \cdot d \]
where \( d \) is the distance between the plates.
- Thus, for \( U_1 \) (between plates 1 and 2):
\[ U_1 = \frac{1}{2} \left( \frac{Q}{\epsilon_0} \right) \left( \frac{3Q}{A \epsilon_0} \cdot d_1 \right)^2 \]
- For \( U_2 \) (between plates 2 and 3):
\[ U_2 = \frac{1}{2} \left( \frac{2Q}{\epsilon_0} \right) \left( \frac{5Q}{A \epsilon_0} \cdot d_2 \right)^2 \]

4. **Find the Ratio \( \frac{U_2}{U_1} \)**:
- The ratio of the potential energies can be simplified:
\[ \frac{U_2}{U_1} = \frac{\frac{1}{2} \left( \frac{2Q}{\epsilon_0} \right) \left( \frac{5Q}{A \epsilon_0} \cdot d_2 \right)^2}{\frac{1}{2} \left( \frac{Q}{\epsilon_0} \right) \left( \frac{3Q}{A \epsilon_0} \cdot d_1 \right)^2} \]
- Canceling out common terms and simplifying gives:
\[ \frac{U_2}{U_1} = \frac{2 \cdot 25 d_2^2}{1 \cdot 9 d_1^2} = \frac{50 d_2^2}{9 d_1^2} \]

5. **Conclusion**:
- The final ratio of the electrostatic potential energies \( \frac{U_2}{U_1} \) depends on the distances \( d_1 \) and \( d_2 \) between the plates.