Question

Two particles A and B having charges 8*10^-6 & -2*10^-6 respectively held fixed with a separation of 20 cm.Where should a third charged particle be placed so that it does not experience a net electric force?

Answer 1

As you know, Coulomb's Law determines the force on the charges. We need a situation such that: \[\left| F_{CA} \right|=\left| F_{CB} \right|\]where the magnitude of force exerted by charge A on charge C is equal to the magnitude of the charge exerted on C by charge B. This becomes:\[\frac{k\left| q_{C}q_{A} \right|}{r_{CA}^2}=\frac{k \left| q_{C}q{B} \right|}{r_{CB}^2}\]rearranging,\[\frac{\left| q_{C}q_{A} \right|}{\left| q_{C}q_{B} \right|}=\frac{r_{CA}^2}{r_{CB}^2}\]\[\frac{\left| q_A \right|}{\left| q_B \right|}=\frac{r_{CA}^2}{r_{CB}^2}\]So,\[\frac{r_{CA}^2}{r_{CB}^2}=4\]And,\[r_{CA}=2r_{CB}\]Now we need to relate the distance from charge A to C to the distance from charge C to B. We do this using the information that the distance from A to B is 20cm. We know that charge C will lie on the lie joining the centres of charges A and B. Putting charge A at the origin and charge B on the x-axis where x=20, charge C will lie in between them. We have,\[r_{AB}=r_{AC}+r_{CB}\] so,\[r_{CB}=20-r_{AC}\] Using this above we find,\[r_{AC}=2(20-r_{AC})\]Thus,\[r_{AC}=\frac{40}{3}\]The third charge should be place in between charges A and B, 40/3 cm's away from charge A and 20/3 cm's aways from charge C to feel zero net electrical force.