Answer to Question #136968 in Electricity and Magnetism for jeff

Question

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 6.89 nC, and the other has a charge of 4.18 nC.
(a)
What is the magnitude (in N) of the electric force that one particle exerts on the other?

Answer 1

Here,

Q₁ = 6.89nC

Q₂ = 4.18nC

r₁₂ = 1.82m

Force exerted by Q₂ on Q₁ is given by-

F₁₂ = $\frac{kQ_1Q_2}{r_{12}^2}$

Here, k = $\frac{1}{4\pi \epsilon_0}$ = 9$\times$ 10⁹

So,

F₁₂ = $\frac{(9\times10^9)(6.89\times10^{-9})(4.18\times10^{-9})}{(1.82)^2}$

= 7.825 $\times10^{-8}$ N

Since both charges are negative, the force between two charges will be repulsive in nature.