Question

A negative charge, q1, of 6 µC is 0.002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

magnitude: 8 × 101 N
direction: south
magnitude: 8 × 101 N
direction: north
magnitude: 4 × 104 N
direction: south
magnitude: 4 × 104 N
direction: north

Answer 1

To find the magnitude and direction of the electrical force applied by the negative charge \( q_1 \) on the positive charge \( q_2 \), we can use Coulomb's law, which is given by the formula:

\[
F_e = k \frac{|q_1 q_2|}{r^2}
\]

where:
- \( F_e \) is the magnitude of the electrical force,
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

Given:
- \( q_1 = -6 \, \mu C = -6 \times 10^{-6} \, C \),
- \( q_2 = 3 \, \mu C = 3 \times 10^{-6} \, C \),
- \( r = 0.002 \, m \).

Now substituting the values into the formula:

\[
F_e = (8.99 \times 10^9) \frac{|(-6 \times 10^{-6})(3 \times 10^{-6})|}{(0.002)^2}
\]

Calculate \( |q_1 q_2| \):

\[
|(-6 \times 10^{-6})(3 \times 10^{-6})| = 18 \times 10^{-12} \, C^2
\]

Calculate \( r^2 \):

\[
(0.002)^2 = 4 \times 10^{-6} \, m^2
\]

Now plug these into the equation:

\[
F_e = (8.99 \times 10^9) \frac{18 \times 10^{-12}}{4 \times 10^{-6}}
\]

\[
F_e = (8.99 \times 10^9) \frac{18}{4} \times 10^{-6}
\]

\[
F_e = (8.99 \times 10^9) \times 4.5 \times 10^{-6}
\]

Now calculating:

\[
F_e \approx 8.99 \times 4.5 \times 10^3
\]

\[
F_e \approx 40.455 \times 10^3 \, N
\]
\[
F_e \approx 4.0455 \times 10^4 \, N \approx 4 \times 10^4 \, N
\]

Now for the direction: The negative charge \( q_1 \) attracts the positive charge \( q_2 \). Since \( q_2 \) is to the south of \( q_1 \), the direction of the force on \( q_2 \) will be towards \( q_1 \), which is downwards (south).

Thus, the final answer is:

**Magnitude:** \( 4 \times 10^{4} \, N \)

**Direction:** South.