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