Question

Two particles with electric charges Q and -3Q are separated by a distance of 1.2 m.

(a) If Q = 4.5C, what is the electric force between the two particles?

(b) If Q= -4.5 C, how does the answer change? 

Answer 1

(1)

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{k(Q)\times(-3Q)}{1.2^2}$

$Q=4.5C$

$$F=\frac{9\times10^9(4.5)\times(-3\times4.5)}{1.2^2}=-3.79\times10^{11}N$$

Attractive force

Part (b)

$F=\frac{kq_1q_2}{r^2}$

$F=\frac{k(Q)\times(-3Q)}{1.2^2}$

$Q=-4.5C$

$$F=\frac{9\times10^9(-4.5)\times(-3\times(-4.5))}{1.2^2}=-3.79\times10^{11}N$$

Attractive force

Answer doesn't change