Certainly! Let's solve this step-by-step.
1. Understand the given data:
- We have an electric field \(\( E \)\) with a magnitude of \(\( 20 \, \text{N/C} \)\).
- There's a point charge \(\( Q \)\) at a certain distance \(\( r \)\) from the source charge.
- The distance given from the charge Q to the point is \(\( 2 \, \text{m} \)\).
- There is a test charge \(\( q \)\) of \(\( 5 \, \text{C} \)\) placed in this electric field.
2. Objective:
- We need to calculate the force \(\( F \)\) exerted by the electric field on the test charge \(\( q \)\).
3. Use the relationship between force, electric field, and charge:
- The relationship between the electric field \(\( E \)\), the force \(\( F \)\), and the charge \(\( q \)\) is given by the formula:
\(\[
F = q \cdot E
\]\)
4. Substitute the known values into the formula:
- The electric field \(\( E \)\) is \(\( 20 \, \text{N/C} \)\).
- The test charge \(\( q \)\) is \(\( 5 \, \text{C} \)\).
\(\[
F = 5 \, \text{C} \times 20 \, \text{N/C}
\]\)
5. Calculate the force:
\(\[
F = 100 \, \text{N}
\]\)
Therefore, the force exerted by the electric field on a charge of \(\( 5 \, \text{C} \)\) is \(\( 100 \, \text{N} \)\).
