- What is the electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m?
- How will this force change if the distance is increased to 1 m?

- $3.60 \times 10^{10} \textrm{ N}$
- $8.99 \times 10^9 \textrm{ N}$

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This is College Physics Answers with Shaun Dychko. The electrostatic force between two charges is Coulomb's constant times the first charge times the second charge divided by the distance between them squared. So that's 8.988 times ten to the nine Newtons meter squared per Coulomb squared, times one Coulomb from the first charge, and times one coulomb from the second charge, divided by 0.5 meters between them squared, giving 3.60 times ten to the ten Newtons of force between them. Now if the distance is increased to one meter, that represents a difference by a factor of two. So we can say that force two is <i>k q one q two</i> over this new distance which is two times the original distance. We square that and this makes a factor of one quarter times <i>F one</i> because that's what all this stuff is. This is the first force. So that's one quarter times 3.5952 times ten to the ten Newtons, writing this number with more digits because we want to avoid intermediate rounding error. We get an answer for the force in part B will be 8.99 times ten to the nine Newtons.