Question

Three positive point charges of 3.0 nC, 6.0 nC, and 2.0 nC,respectively, are arranged in a triangular pattern, as shown. Findthe magnitude and direction of theelectrical force acting on the6.0 nC charge.

Answer 1

The three positive point charges are q₁= 3.0nC,q₂= 6.0 nC and q₃= 2.0 nCThe electrical force acting on the charge q₂ due tothe charge q₁ isF₁= (1/4πε_o) *(q₁q₂/r²)Here,(1/4πε_o) = 9 * 10⁹Nm²/C² and r = (1 + 1)^1/2 =√2or F₁=9 * 10⁹* (3.0 *10^-9* 6.0 *10^-9/(√2)²)or F₁= 81 * 10^-9 NThe electrical force acting on the charge q₂ due tothe charge q₃ isF₂= 9 * 10⁹* (6.0 *10^-9* 2.0 *10^-9/(√2)²)or F₂=54 * 10^-9 NThe magnitude of the net electrical force acting on the 6.0 nCcharge isF = F₁ + F₂or F = 81 * 10^-9 +54 * 10^-9Nor F =135 * 10^-9 NThe direction of the net electrical force acting on the 6.0 nCcharge istanα = (F₁ * cosθ/F₁ +F₂ * sinθ)Here,sinθ = (1/√2)or θ = 45^oor tanα = (81 * 10^-9 * cos(45^o)/81* 10^-9 +54 * 10^-9 *sin(45^o))or tanα = 0.4807or α = 25.67^oTherefore,the direction of the net electrical force acting onthe 6.0 nC charge is α = 25.67^o counterclockwisefrom the positive x-axis.