1. **Problem Statement:** A test charge of +3 µC is at a point P where the electric field due to other charges is directed to the right and has a magnitude of $4 \times 10^6$ N/C. If the test charge is replaced with a charge of -3 µC, what happens to the electric field at P? 2. **Relevant Formula:** The electric field $\vec{E}$ at a point is defined as the force $\vec{F}$ on a test charge $q_0$ divided by the magnitude of that test charge: $$\vec{E} = \frac{\vec{F}}{q_0}$$ 3. **Explanation:** - The electric field $\vec{E}$ depends only on the source charges creating the field, not on the test charge. - Changing the test charge from +3 µC to -3 µC changes the direction of the force $\vec{F}$ on the test charge because force direction depends on the sign of $q_0$. - However, the electric field $\vec{E}$ itself remains the same in magnitude and direction because it is a property of the source charges. 4. **Answer:** - The electric field at point P has the **same magnitude** as before. - The direction of the electric field does **not** change when the test charge changes. - The force on the test charge changes direction because the test charge sign changed. Therefore, the correct choice is **(c) remains the same**. **Final answer:** The electric field at P remains the same in magnitude and direction when the test charge is replaced with -3 µC.