What must be a distance (in meters) between a point charge of -5 times 10^{-7} C and another...


What must be a distance (in meters) between a point charge of {eq}-5 \times 10^{-7}\ C {/eq} and another charge of {eq}+3 \times 10^{-7}\ C {/eq} so that the magnitude of the force between them will be equal to {eq}0.058\ N {/eq}? {eq}(k = 9 \times 10^9\ N \cdot m^2 / C^2) {/eq}

Electrostatic Force between the two stationary charges:

The electrostatic force is defined as the force acting between the stationary charged particles. According to Coulomb's law, the electrostatic force is expressed as such;

{eq}F_{C} \propto q_{1} q_{2} \,\,\,\,\,\, {/eq}, {eq}F_{C} \propto \dfrac{1}{r^{2}} {/eq}


{eq}F_{C} = \dfrac{kq_{1}q_{2}}{r^{2}} {/eq}


  • k is the electrostatic force constant.
  • r is the separation between two stationary charges.


  • The electrostatic force between two charged particles or objects is either attractive or repulsive in nature. It depends on the polarity of the stationary charges.

Answer and Explanation:

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We are given the following information:

{eq}q_{1} = - 5.0 \times 10^{-7} \, C. {/eq}

  • {eq}q_{2} =+3.0 \times 10^{-7} \, C. {/eq}
  • The electrostatic...

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Learn more about this topic:

Coulomb's Law: Variables Affecting the Force Between Two Charged Particles


Chapter 6 / Lesson 3

In the 18th century, Charles Coulomb uncovered the secrets of electrostatic force between charged particles. The results of his experiments led to what is now known as Coulomb's Law, which tells us how force, charge, and distance are all related.

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