Question

Particles q1 = -53.0 uC, q2= +105 uC, and q3= -88.0 uC are in a line. Particles q1 and q2 are separated by 0.50m and particles q2 and q3 are separated by 0.95m. What is the net force on particle q3?

Answer 1

Here $r_{12} = 0.5 m$, $r_{23} = 0.95m$. Then the distance from the first charge to the third one will be:

$$r_{13} = r_{12} + r_{23} = 0.5 + 0.95 = 1.45 m$$

.

According to the Coloumb's law, the net force on particle q3 will be equal to the sum of the focres on particle q3 from the particle q1 and q2.

$$F_3 = F_{13} + F_{23} = k\dfrac{q_1q_3}{r_{13}^2} + k\dfrac{q_2q_3}{r_{23}^2}$$

where $k = 9\times 10^9 N\cdot m^2/C^2$.

Substituting numerical values, obtain:

$$F_3 = 9\times 10^9 \cdot \dfrac{(-53\times 10^{-6})\cdot (-88\times 10^{-6})}{1.45^2} +\\ + 9\times 10^9 \cdot \dfrac{105\times 10^{-6}\cdot (-88\times 10^{-6})}{0.95^2} \approx \\ \approx-72.18N$$

Answer. 72.18 N.