Question

The attractive electrostatic force between the point charges +8.46 ✕ 10-6 and q has a magnitude of 0.967 n when the separation between the charges is 0.52 m. find the sign and magnitude of the charge q. c

Answer 1

The electrostatic force between two charges is given by

`F=k_e (Q q)/(r^2)`
where

`k_e = 8.99 \cdot 10^9 N m^2 C^(-2)` is the Coulomb's constant

`Q=+8.46 \cdot 10^(-6) C` is one of the charges
q is the other charge
r=0.52 m is the distance between the two charges

We know the intensity of the force between the two charges, F=0.967 N, so we can re-arrange the formula to find the value of the charge q:

`q= (F r^2)/(k_e Q)= ((0.967 N)(0.52 m)^2)/((8.99 \cdot 10^9 N m^2 C^(-2))(+8.46 \cdot 10^(-6) C)) =3.44 \cdot 10^(-6) C`

And the sign of the charge is positive, because Q is positive and F is positive as well.