Two positive point charges are separated by a distance R. If the distance between the charges is reduced to R/2, what happens to the total electric potential energy of the system?

Options

A. It is doubled
B. It is reduced to one-half of its original value
C. It remains the same
D. It is reduced to one-fourth of its original value

Answer

To solve the problem of how the total electric potential energy of a system of two positive point charges changes when the distance between them is reduced, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for electric potential energy (U)**:
The electric potential energy (U) between two point charges (Q1 and Q2) separated by a distance (r) is given by the formula:
\[ U = k \frac{Q_1 Q_2}{r} \]
where \( k \) is Coulomb's constant.

2. **Identify the initial conditions**:
Let both charges be equal, say \( Q \). The initial distance between the charges is \( R \). Therefore, the initial potential energy \( U \) can be expressed as:
\[ U = k \frac{Q \cdot Q}{R} = k \frac{Q^2}{R} \]

3. **Change the distance**:
Now, the distance between the charges is reduced to \( \frac{R}{2} \). We need to find the new potential energy \( U' \) when the distance is \( \frac{R}{2} \):
\[ U' = k \frac{Q \cdot Q}{\frac{R}{2}} = k \frac{Q^2}{\frac{R}{2}} = k \frac{2Q^2}{R} \]

4. **Relate the new potential energy to the initial potential energy**:
From the expressions for \( U \) and \( U' \):
- Initial potential energy: \( U = k \frac{Q^2}{R} \)
- New potential energy: \( U' = k \frac{2Q^2}{R} = 2 \left( k \frac{Q^2}{R} \right) = 2U \)

5. **Conclusion**:
The total electric potential energy of the system when the distance is reduced to \( \frac{R}{2} \) is doubled. Therefore, the final answer is:
\[ U' = 2U \]

### Final Answer:
The total electric potential energy of the system is doubled when the distance between the charges is reduced to \( \frac{R}{2} \).