Question

Object A has a negative charge of -0.25 C. Object B has a positive charge of 1 C. If the distance between A and B is 1.5 m, what is the force on A? Note: the constant of proportionality is 9.0 x 10^9 N • m^2 / C^2.

Group of answer choices

-3.5 x 109 N

6.5 x 109 N

3.5 x 109 N

-1.0 x 109 N

Answer 1

To calculate the electrostatic force between two charged objects, we can use Coulomb's Law, which is given by the formula:

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

where:
- \( F \) is the magnitude of the force between the charges,
- \( k \) is Coulomb's constant (\( 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the amounts of the charges, and
- \( r \) is the distance between the charges.

Given:
- Charge of Object A, \( q_A = -0.25 \, \text{C} \)
- Charge of Object B, \( q_B = 1.0 \, \text{C} \)
- Distance \( r = 1.5 \, \text{m} \)

First, we will compute the magnitude of the force:

\[
F = k \frac{|q_A \cdot q_B|}{r^2} = 9.0 \times 10^9 \frac{|(-0.25) \cdot (1.0)|}{(1.5)^2}
\]

Calculating the denominator:

\[
(1.5)^2 = 2.25
\]

Now substituting into the formula:

\[
F = 9.0 \times 10^9 \frac{0.25}{2.25}
\]

Calculating \( \frac{0.25}{2.25} \):

\[
\frac{0.25}{2.25} \approx 0.1111
\]

Now substituting that back into the force equation:

\[
F \approx 9.0 \times 10^9 \times 0.1111 \approx 1.0 \times 10^9 \, \text{N}
\]

Now we need to take into account the signs of the charges. Object A has a negative charge and Object B has a positive charge, so they attract each other, which means the force on object A will be directed towards object B (and thus will be negative in a conventional sense).

Hence, the force on A will be:

\[
F \approx -1.0 \times 10^9 \, \text{N}
\]

The correct answer, considering the direction of the force, is:

**-1.0 x 10^9 N**