Question

Two point charges of 9uC and 4uC are placed a distance of 4 m
apart.
Calculate the magnitude of the force, in Newtons, on a positive
5uC charge at the midpoint between the two charges.

Answer 1

Answer:

The magnitude of force on the charge at the midpoint is 0.1463N.

Explanation:

Given;

two point charges, q₁ and q₂ = 9μC and 4μC

a positive charge placed at the midpoint between the two charges, q = 5μC

The repulsive force between 9μC and 5μC is calculated by Coulomb's force;

$f_1 = \frac{kq_1q}{r_1^2}$

The repulsive force between 4μC and 5μC is calculated by Coulomb's force;

$f_2 = \frac{kq_2q}{r_2^2}$

The resultant force on the charge is calculated as;

$F= f_1 +f_2 = \frac{kq_1q}{r_1^2} = \frac{kq_2q}{r_2^2} \\\\F = \frac{(9\times 10^9)(9\times 10^{-6})(5\times 10^{-6})}{(2)^2} + \frac{(9\times 10^9)(4\times 10^{-6})(5\times 10^{-6})}{(2)^2} \\\\F = 0.1013 \ N \ + \ 0.045 \ N\\\\F = 0.1463 \ N$

Therefore, the magnitude of force on the charge at the midpoint is 0.1463N.