Question

Object A has a positive charge of 2 C. Object B has a positive charge of 3 C. If the distance between A and B is 0.5 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

5.13 x 1011 N

4.11 x 1011 N

2.16 x 1011 N

9 x 1011 N

Answer 1

To find 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 force between the charges,
- \( k \) is Coulomb's constant (\( k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the centers of the two charges.

Given:
- \( q_1 = 2 \, \text{C} \)
- \( q_2 = 3 \, \text{C} \)
- \( r = 0.5 \, \text{m} \)

Now we can substitute the values into the formula:

\[
F = 9.0 \times 10^9 \frac{|2 \cdot 3|}{(0.5)^2}
\]

Calculating \( (0.5)^2 \):

\[
(0.5)^2 = 0.25
\]

Now substituting this back into the formula:

\[
F = 9.0 \times 10^9 \frac{6}{0.25}
\]

Calculating \( \frac{6}{0.25} \):

\[
\frac{6}{0.25} = 6 \times 4 = 24
\]

Now substituting this value back into the equation for \( F \):

\[
F = 9.0 \times 10^9 \times 24
\]

Calculating \( 9.0 \times 10^9 \times 24 \):

\[
F = 216 \times 10^9 = 2.16 \times 10^{11} \, \text{N}
\]

Thus, the force on object A is:

\[
F = 2.16 \times 10^{11} \, \text{N}
\]

The correct answer is:

**2.16 x 10^11 N**