Question

Point charges q₁ = +21μC and q₂ = −50μC are placed 2.59 m apart. What is the force on a third charge q₃ = 18μC placed midway between q₁ and q₂?

Round your answer to 2 decimal places

Answer 1

Accoridng to the superposition principle, the force on the third charge is:

$$F_3 = F_{31} + F_{32}$$

where

$$F_{31} = k\dfrac{|q_1||q_3|}{r_{31}^2}, \space\space\space F_{32} = k\dfrac{|q_2||q_3|}{r_{32}^2}$$

is the Coulomb's force on the third charge due to the first and second charges respectively.

Here $k = 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant, $r_{31} = r_{32} = 2.59m/2 = 1.295m$ are the distances from the first and second charges to the third charge respectively. Thus, obtain:

$$F_3 = \dfrac{k|q_3|}{r_{31}^2}(|q_1| + |q_2|)\\ F_3 = \dfrac{ 9\times 10^9\cdot18\times 10^{-6}}{1.295^2}(21\times 10^{-6}+ 50\times 10^{-6}) \approx 6.9N$$

Answer. 6.9 N.