Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an identical charge from infinity to the third vertex is

Answer

To solve the problem of calculating the work done in bringing an identical charge from infinity to the third vertex of an equilateral triangle where two identical charges are already placed at the other two vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have an equilateral triangle with two identical charges \( Q \) placed at two corners (let's call them \( A \) and \( B \)).
- The side length of the triangle is \( A \).
- We need to bring a third identical charge \( Q \) from infinity to the third corner (let's call it \( C \)).

2. **Calculate Initial Potential Energy**:
- The potential energy \( U \) of the system with the two charges \( Q \) at points \( A \) and \( B \) can be calculated using the formula for potential energy between two point charges:
\[ U = \frac{k Q^2}{A} \]
- Here, \( k \) is Coulomb's constant.

3. **Calculate Final Potential Energy**:
- When we bring the third charge \( Q \) to point \( C \), we need to consider the potential energy of the new configuration.
- The final potential energy \( U_f \) will include the interactions between all three charges:
- The potential energy between charge at \( A \) and charge at \( C \):
\[ U_{AC} = \frac{k Q^2}{A} \]
- The potential energy between charge at \( B \) and charge at \( C \):
\[ U_{BC} = \frac{k Q^2}{A} \]
- The potential energy between charges at \( A \) and \( B \) (which we already calculated):
\[ U_{AB} = \frac{k Q^2}{A} \]
- Therefore, the total final potential energy \( U_f \) is:
\[ U_f = U_{AB} + U_{AC} + U_{BC} = \frac{k Q^2}{A} + \frac{k Q^2}{A} + \frac{k Q^2}{A} = 3 \frac{k Q^2}{A} \]

4. **Calculate Work Done**:
- The work done \( W \) in bringing the charge from infinity to point \( C \) is equal to the change in potential energy:
\[ W = U_f - U \]
- Substituting the values we have:
\[ W = 3 \frac{k Q^2}{A} - \frac{k Q^2}{A} = 2 \frac{k Q^2}{A} \]
- Since \( U = \frac{k Q^2}{A} \), we can express the work done in terms of \( U \):
\[ W = 2U \]

### Final Answer:
The work done in bringing the identical charge from infinity to the third vertex is \( 2U \).