To solve the problem, we need to calculate the change in potential energy when one charge is moved to the midpoint of the line joining the other two charges, and then determine how long it would take to supply that energy at a rate of 1 kW.
### Step-by-Step Solution:
1. **Identify the Charges and Configuration**:
We have three charges, each of `0.1 C`, placed at the corners of an equilateral triangle with a side length of `1 m`.
2. **Calculate Initial Potential Energy**:
The potential energy \( U_i \) of the system when all three charges are at the corners can be calculated using the formula for potential energy between point charges:
\[
U_i = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right)
\]
Here, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( q_1 = q_2 = q_3 = 0.1 \, C \), and \( r_{12} = r_{13} = r_{23} = 1 \, m \).
Thus,
\[
U_i = k \left( \frac{0.1 \times 0.1}{1} + \frac{0.1 \times 0.1}{1} + \frac{0.1 \times 0.1}{1} \right) = 3 \times k \times \frac{0.01}{1} = 3 \times 9 \times 10^9 \times 0.01 = 2.7 \times 10^8 \, J
\]
3. **Calculate Final Potential Energy**:
When one charge is moved to the midpoint of the line joining the other two charges, the distances change. The distance between the two charges remains `1 m`, but the distance from the midpoint to each charge is `0.5 m`.
The potential energy \( U_f \) in this new configuration is:
\[
U_f = k \left( \frac{q_1 q_2}{0.5} + \frac{q_1 q_3}{0.5} + \frac{q_2 q_3}{1} \right)
\]
\[
U_f = k \left( \frac{0.1 \times 0.1}{0.5} + \frac{0.1 \times 0.1}{0.5} + \frac{0.1 \times 0.1}{1} \right) = k \left( 2 \times \frac{0.01}{0.5} + \frac{0.01}{1} \right)
\]
\[
U_f = k \left( 2 \times 0.02 + 0.01 \right) = k \times 0.05 = 9 \times 10^9 \times 0.05 = 4.5 \times 10^8 \, J
\]
4. **Calculate Change in Potential Energy**:
The change in potential energy \( \Delta U \) is given by:
\[
\Delta U = U_f - U_i = 4.5 \times 10^8 - 2.7 \times 10^8 = 1.8 \times 10^8 \, J
\]
5. **Calculate Time Required**:
The energy is supplied at a rate of \( 1 \, kW = 1000 \, J/s \). The time \( t \) required to supply the energy \( \Delta U \) is:
\[
t = \frac{\Delta U}{\text{Power}} = \frac{1.8 \times 10^8}{1000} = 1.8 \times 10^5 \, s
\]
### Final Answer:
The time required to move one of the charges to the midpoint of the line joining the other two is \( 1.8 \times 10^5 \, s \).
