Question

Particles q1 = -53.0 uc, q2 = +105 uc, and

Answer 1

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$

The force on the particle $q_1$ is $-180.38\ \text{N}$.