Electrostatics Part 2

Charge of particle $A, q_{A}=20\ \mu\mathrm{C}$ Charge of particle $B, q_{B}=-5\ \mu\mathrm{C}$ Separation between A and $B, R=5\ \mathrm{cm}=5 \times 10^{-2}$ The given situation is shown below. Here, $q_{A}=20\ \mu\mathrm{C}, q_{B}=-5\ \mu\mathrm{C}, r=5\ \mathrm{cm}$ Let, third charge particle of charge $q_{C}$ is placed at point C to the right of charge B, so that net electric force would be equal to zero. This is because of following reason. As the net electric force on particle C should be equal to zero, so the forces due to charged particles A and B must be in opposite direction. Hence, the particle should be placed on the line AB. As particles A and B having charges of opposite signs, so charged particle C cannot be between A and B. Also, A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. Now, let position of C is at distance x from right of B. ∴ Force on C due to $B(F_{CB})=$ Force on C due to $A(F_{CA})$ $\Rightarrow \frac{k q_{B} q_{C}}{x^{2}} = \frac{k q_{A} q_{C}}{\left(x+\frac{5}{100}\right)^{2}}$ $\Rightarrow \frac{\left(x+\frac{5}{100}\right)^{2}}{x^{2}} = \frac{q_{A}}{q_{B}}$ $\Rightarrow \frac{20}{5} = \left(\frac{x+\frac{5}{100}}{x}\right)^{2}$ $\Rightarrow 2 = \frac{x+\frac{5}{100}}{x}$ $\Rightarrow 2x = x + \frac{5}{100}$ ⇒ x = 5 cm