Question

The attractive electrostatic force between the point charges 8.11×10⁻⁶ C and Q has a magnitude of 0.495 N when the separation between the charges is 5.40 m.

find the sign and magnitude of the charge Q

Answer 1

From Coulomb's inverse square law the force of attraction or repulsion is given by $F= \frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{R^{2}}$

Here in our problem 0.495 = 9x10⁹x8.22x10^-6 xQ/(5.4)²

Therefore the charge on the second body is

                                 Q = 0.495x(5.4)²/ (9x10⁹x8.11x10^-6)

                                  Q = 198x10^-3C

Since the force is attractive the charges should be opposite in charge. Hence this charge should be negative.

                                Therefore Q=-198x10^-3C