Certainly! Let's break down the problem step-by-step.
### Part 1: Calculating the Electric Field (E)
We are asked to calculate the strength of the electric field (E) generated by the point charge.
#### Given:
- The point charge [tex]\( q \)[/tex] is [tex]\( 2.0 \, \text{nC} \)[/tex] (nanocoulombs).
- The distance [tex]\( r \)[/tex] from the charge is [tex]\( 5.0 \, \text{mm} \)[/tex] (millimeters).
- Coulomb's constant [tex]\( k \)[/tex] is [tex]\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)[/tex].
First, convert the given units into standard SI units:
- [tex]\( 2.0 \, \text{nC} = 2.0 \times 10^{-9} \, \text{C} \)[/tex].
- [tex]\( 5.0 \, \text{mm} = 5.0 \times 10^{-3} \, \text{m} \)[/tex].
The formula for the electric field (E) created by a point charge is given by:
[tex]\[ E = \frac{k \cdot q}{r^2} \][/tex]
Substituting the given values:
[tex]\[ E = \frac{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \times 2.0 \times 10^{-9} \, \text{C}}{(5.0 \times 10^{-3} \, \text{m})^2} \][/tex]
After performing the calculation:
[tex]\[ E \approx 719200 \, \text{N/C} \][/tex]
This is the strength of the electric field created by the point charge at the given distance.
### Part 2: Calculating the Force on the Test Charge
We now need to calculate the force exerted by this electric field on a test charge.
#### Given:
- The test charge [tex]\( q_{\text{test}} \)[/tex] is [tex]\( -0.25 \, \mu\text{C} \)[/tex] (microcoulombs).
Convert this to standard SI units:
- [tex]\( -0.25 \, \mu\text{C} = -0.25 \times 10^{-6} \, \text{C} \)[/tex].
The force [tex]\( F \)[/tex] exerted on a charge in an electric field is given by:
[tex]\[ F = q_{\text{test}} \cdot E \][/tex]
Substituting the given values:
[tex]\[ F = (-0.25 \times 10^{-6} \, \text{C}) \times 719200 \, \text{N/C} \][/tex]
After performing the calculation:
[tex]\[ F \approx -0.1798 \, \text{N} \][/tex]
### Conclusion
- The strength of the electric field [tex]\( E \)[/tex] at a distance of [tex]\( 5.0 \, \text{mm} \)[/tex] from a [tex]\( 2.0 \, \text{nC} \)[/tex] point charge is approximately [tex]\( 719200 \, \text{N/C} \)[/tex].
- The force [tex]\( F \)[/tex] exerted by this electric field on a [tex]\( -0.25 \, \mu\text{C} \)[/tex] test charge is approximately [tex]\( -0.1798 \, \text{N} \)[/tex].
The negative sign in the force indicates that the force is attractive (i.e., the direction of the force is towards the point charge, since the test charge is negative and the point charge is positive).
