Question

Three balls, with charges of +a, -2q, and +3q, are placed at the corners of a square measuring L on each side, as shown in the figure. The value of q is 3.20 x 10-6 C, and L = 40.0 cm. Assume this set of three balls is not interacting with anything else in the universe, and assume that gravitational interactions are negligible -2q t9 L. +3q (a) What is the magnitude of the net electrostatic force on the ball of charge-2q 3.643 (b) What is the magnitude of the net electrostatic force on the ball of charge +3q?

Answer 1

Force due to +q charge

F1 = k q (3q) / (2 L^2)

F1 = 9*10^9* 3*(3.2*10^-6)^2 / (2* 0.4^2)

F1 = 0.864 N

Force due to - 2q charge

F2 = k 3q 2q / L^2

F2 = 9*10^9* 6* (3.2*10^-6)^2 / 0.4^2

F2 = 3.456 N

Angle between the forces

X = 45

Now Resultant magnitude of force acting on q3

F^2 = F1^2 + F2^2 + 2 F1 F2 cos x

F = 4.11 N

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