What is the magnitude and direction of the electric force on a `-1.2 mu C` charge at a point where the electric field is 2500 N/C and is directed along the +y axis.

Options

A. 0.15 N, -y direction
B. 0.0030 N, -y direction
C. 0.15 N, +y direction
D. 0.0030 N, +y direction

Answer

To find the magnitude and direction of the electric force on a charge in an electric field, we can use the formula:

\[ F = Q \times E \]

where:
- \( F \) is the electric force,
- \( Q \) is the charge,
- \( E \) is the electric field strength.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Charge \( Q = -1.2 \, \mu C = -1.2 \times 10^{-6} \, C \)
- Electric field \( E = 2500 \, N/C \) (directed along the +y axis)

2. **Calculate the Magnitude of the Electric Force**:
- Using the formula \( F = Q \times E \):
\[ F = (-1.2 \times 10^{-6} \, C) \times (2500 \, N/C) \]
- Calculate the product:
\[ F = -3.0 \times 10^{-3} \, N \]
- The magnitude of the force is \( 3.0 \times 10^{-3} \, N \) or \( 0.003 \, N \).

3. **Determine the Direction of the Electric Force**:
- The electric field is directed along the +y axis. For a positive charge, the force would also be in the +y direction. However, since the charge is negative, the force will be in the opposite direction.
- Therefore, the direction of the force is along the -y axis (or negative j direction).

### Final Answer:
- **Magnitude of the Electric Force**: \( 0.003 \, N \)
- **Direction of the Electric Force**: Along the -y axis (negative j direction).

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