A point charge `q_1 =-5.8muC` is held stationary at the origin. A second point charge `q_2=+4.3muC` moves from the point `(0.26m 0,0)` to `(0.38m, 0, 0)`. How much work is done by the electric force on `q_2`?

Explanation

To find the work done by the electric force on the charge \( q_2 \) as it moves from its initial position to its final position, we can use the concept of electric potential energy. The work done by the electric force is equal to the negative change in potential energy.

### Step-by-Step Solution:

1. **Identify the Charges and Positions:**
- The stationary charge \( q_1 = -5.8 \, \mu C = -5.8 \times 10^{-6} \, C \) is located at the origin (0, 0, 0).
- The moving charge \( q_2 = +4.3 \, \mu C = 4.3 \times 10^{-6} \, C \) moves from the initial position \( (0.26 \, m, 0, 0) \) to the final position \( (0.38 \, m, 0, 0) \).

2. **Calculate the Initial and Final Distances:**
- The initial distance \( r_i \) from \( q_1 \) to \( q_2 \) is:
\[ r_i = 0.26 \, m \]
- The final distance \( r_f \) from \( q_1 \) to \( q_2 \) is:
\[ r_f = 0.38 \, m \]

3. **Calculate the Electric Potential Energy:**
- The electric potential energy \( U \) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
- Where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \).

4. **Calculate Initial Potential Energy \( U_i \):**
\[ U_i = k \frac{q_1 q_2}{r_i} = 9 \times 10^9 \cdot \frac{(-5.8 \times 10^{-6})(4.3 \times 10^{-6})}{0.26} \]

5. **Calculate Final Potential Energy \( U_f \):**
\[ U_f = k \frac{q_1 q_2}{r_f} = 9 \times 10^9 \cdot \frac{(-5.8 \times 10^{-6})(4.3 \times 10^{-6})}{0.38} \]

6. **Calculate the Work Done \( W \):**
- The work done by the electric force is given by:
\[ W = - (U_f - U_i) = U_i - U_f \]

7. **Substituting Values:**
- Calculate \( U_i \) and \( U_f \):
\[ U_i = 9 \times 10^9 \cdot \frac{(-5.8 \times 10^{-6})(4.3 \times 10^{-6})}{0.26} \approx -0.272 \, J \]
\[ U_f = 9 \times 10^9 \cdot \frac{(-5.8 \times 10^{-6})(4.3 \times 10^{-6})}{0.38} \approx -0.187 \, J \]
- Now calculate \( W \):
\[ W = - (-0.187 - (-0.272)) = 0.272 \, J \]

### Final Answer:
The work done by the electric force on \( q_2 \) is approximately \( 0.272 \, J \).

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