Gravitational potential at point 'P' due to triangular plate is `V_0` . Find the gravitational potential due to the given plate at point 'Q' as shown in figure. (both plates are made by same meterial)

Options

A. `V_0`
B. `2V_0`
C. `4V_0`
D. `3V_0`

Answer

To find the gravitational potential at point 'Q' due to the triangular plate, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The gravitational potential at point 'P' due to the triangular plate is given as \( V_0 \).
- The triangular plate has sides of length \( A \).
- The distance from point 'Q' to the triangular plate is also \( A \).

2. **Determine the Geometry**:
- The distance from point 'P' to the center of mass of the triangular plate is \( A \).
- The distance from point 'Q' to the center of mass of the triangular plate is \( 2A \) (since it is at a distance of \( A \) from the plate and the plate's height is also \( A \)).

3. **Use the Formula for Gravitational Potential**:
- The gravitational potential \( V \) due to a mass \( M \) at a distance \( r \) is given by:
\[ V \propto \frac{M}{r} \]
- Since the mass \( M \) of the triangular plate is constant, we can express the potentials at points 'P' and 'Q' in terms of their distances from the plate.

4. **Relate the Potentials**:
- At point 'P', the potential is:
\[ V_P \propto \frac{M}{A} \]
- At point 'Q', the potential is:
\[ V_Q \propto \frac{M}{2A} \]

5. **Express \( V_Q \) in Terms of \( V_0 \)**:
- Since \( V_0 \) is defined as the potential at point 'P':
\[ V_0 \propto \frac{M}{A} \]
- Therefore, we can express \( V_Q \) as:
\[ V_Q \propto \frac{M}{2A} = \frac{1}{2} \cdot \frac{M}{A} = \frac{1}{2} V_0 \]

6. **Final Expression**:
- Thus, the gravitational potential at point 'Q' can be expressed as:
\[ V_Q = \frac{1}{2} V_0 \]

### Conclusion:
The gravitational potential at point 'Q' is \( \frac{1}{2} V_0 \).