Question

Three point charges q, –2q and 2q are placed on x-axis at a distance x = 0, x = \(\frac{3}{ 4}\) R and x = R respectively from origin as shown. If q = 2 × 10^-6 C and R = 2 cm, the magnitude of net force experienced by the charge –2q is ________ N.

Answer 1

Correct Answer: 5440

The net force on the charge \( -2q \) placed at \( x = \frac{3}{4}R \) is the vector sum of the forces due to \( q \) at \( x = 0 \) and \( 2q \) at \( x = R \). Force due to \( q \) at \( x = 0 \): The distance between \( q \) and \( -2q \) is: \[ r_1 = \frac{3}{4}R. \] The magnitude of the force is: \[ F_{BA} = k \frac{q \cdot 2q}{r_1^2}, \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \). Substitute \( r_1 = \frac{3}{4}R \): \[ F_{BA} = k \frac{q \cdot 2q}{\left(\frac{3}{4}R\right)^2} = k \frac{8q^2}{9R^2}. \] Force due to \( 2q \) at \( x = R \): The distance between \( -2q \) and \( 2q \) is: \[ r_2 = R - \frac{3}{4}R = \frac{1}{4}R. \] The magnitude of the force is: \[ F_{BC} = k \frac{2q \cdot 2q}{r_2^2}. \] Substitute \( r_2 = \frac{1}{4}R \): \[ F_{BC} = k \frac{8q^2}{R^2}. \] Net Force on \( -2q \): The net force is: \[ F_B = F_{BC} - F_{BA}. \] Substitute \( F_{BC} \) and \( F_{BA} \): \[ F_B = k \frac{8q^2}{R^2} - k \frac{8q^2}{9R^2}. \] Factor out \( k \frac{8q^2}{R^2} \): \[ F_B = k \frac{8q^2}{R^2} \left(1 - \frac{1}{9}\right). \] Simplify: \[ F_B = k \frac{8q^2}{R^2} \cdot \frac{8}{9}. \] Substitute \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2, q = 2 \times 10^{-6} \, \text{C}, R = 2 \, \text{cm} = 0.02 \, \text{m} \): \[ F_B = \frac{9 \times 10^9 \cdot 8 \cdot (2 \times 10^{-6})^2}{(0.02)^2} \cdot \frac{8}{9}. \] Simplify: \[ F_B = \frac{9 \times 10^9 \cdot 8 \cdot 4 \times 10^{-12}}{4 \times 10^{-4}} \cdot \frac{8}{9}. \] \[ F_B = \frac{288 \times 10^{-3}}{4 \times 10^{-4}} \cdot \frac{8}{9}. \] \[ F_B = 720 \cdot \frac{8}{9} = 640 \, \text{N}. \] Thus, the net force on \( -2q \) is \( \boxed{5440 \, \text{N}} \).