Object A has a negative charge of -0.25 C. Object B has a negative charge of -0.75 C. If the distance between A and B is 0.05 m, what is the force on A? Note: the constant of proportionality is 9.0 x 10^9 N • m^2 /C^2.

Question

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Group of answer choices

6.75 x 1011 N

9.75 x 1011 N

3.25 x 1011 N

6.75 x 10-11 N

Answer 1

To find the electrostatic force between objects A and B, 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 (\( 9.0 \times 10^9 , \text{N} \cdot \text{m}^2/\text{C}^2 \)),
• \( q_1 \) and \( q_2 \) are the charges of the objects,
• \( r \) is the distance between the charges.

Given:

• \( q_1 = -0.25 , \text{C} \)
• \( q_2 = -0.75 , \text{C} \)
• \( r = 0.05 , \text{m} \)

Now, substituting these values into the formula:

\[ F = 9.0 \times 10^9 , \frac{|-0.25 \cdot -0.75|}{(0.05)^2} \]

Calculating \( |q_1 q_2| \):

\[ |-0.25 \cdot -0.75| = 0.25 \cdot 0.75 = 0.1875 , \text{C}^2 \]

Now, calculating \( r^2 \):

\[ (0.05)^2 = 0.0025 , \text{m}^2 \]

Now substitute back into the force equation:

\[ F = 9.0 \times 10^9 , \frac{0.1875}{0.0025} \]

Calculating \( \frac{0.1875}{0.0025} \):

\[ \frac{0.1875}{0.0025} = 75 \]

Now calculating \( F \):

\[ F = 9.0 \times 10^9 \times 75 = 6.75 \times 10^{11} , \text{N} \]

The force on object A is:

\[ F = 6.75 \times 10^{11} , \text{N} \]

Thus, the correct answer is:

6.75 x 10^11 N