Questions: When an object with an electric charge of 0.770 mC is 90.0 m from an object with an electric charge of -0.890 mC, the force between them has a strength of 0.7604 N. Calculate the strength of the force between the two objects if they are 450 m apart. Be sure your answer has the correct number of significant digits.

Solution Steps

Coulomb's Law states that the magnitude of the electrostatic force \( F \) between two point charges is given by: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where:

• \( k_e \) is Coulomb's constant, approximately \( 8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges
• \( r \) is the distance between the charges

We are given:

• \( q_1 = 0.770 \, \text{mC} = 0.770 \times 10^{-3} \, \text{C} \)
• \( q_2 = -0.890 \, \text{mC} = -0.890 \times 10^{-3} \, \text{C} \)
• Initial distance \( r_1 = 90.0 \, \text{m} \)
• Initial force \( F_1 = 0.7604 \, \text{N} \)

We need to find the force \( F_2 \) when the distance \( r_2 = 450.0 \, \text{m} \).

Using the inverse square law property of Coulomb's Law: \[ \frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^2 \]

Substituting the given values: \[ \frac{F_2}{0.7604} = \left( \frac{90.0}{450.0} \right)^2 \] \[ \frac{F_2}{0.7604} = \left( \frac{1}{5} \right)^2 \] \[ \frac{F_2}{0.7604} = \frac{1}{25} \] \[ F_2 = 0.7604 \times \frac{1}{25} \] \[ F_2 = 0.7604 \times 0.04 \] \[ F_2 = 0.03042 \, \text{N} \]

Final Answer