Question

Two-point charges are located on the y-axis as follows: charge${\mathit{q}}_{\mathbf{1}}$= -1.50 nC at y = -0.600 m, and charge${\mathit{q}}_{\mathbf{2}}$= +3.20 nC at the origin (y = 0). What is the total force (magnitude and direction) exerted by these two charges on a third charge${\mathit{q}}_{\mathbf{3}}$= +5.00 nC located at y = -0.400 m?

Short answer

The total force exerted by two charges on a third charge is$2.583\times {10}^{-6}N$.

Step-by-step solution

Using Coulomb Law

Using Coulomb’s law;

$\mathbf{F}\mathbf{=}\mathbf{K}\left|\frac{q_1{q}_2}{r^2}\right|$, here the law states that the force of attraction or repulsion between the two charges is directly proportional to the product of charges and inversely proportional to the distance between them.

The forces exerted by charges on each other and if two forces are equal in magnitude and opposite in direction, even when the charges are not equal. The forces obey Newton’s third law.

When two charges exert forces on a third charge simultaneously, the total force acting on that charge is a vector sum of force that two charges are exerting.

Calculation

Here, $F_1\to$force exerted by the first charge on the third charge

$$\begin{gathered} r=y_3-y_1 \\ =-0.4-(-0.6) \\ =0.2m \end{gathered}$$

Force is attractive in the negative y-axis

$$\begin{gathered} F_1=K\frac{\left|q_1{q}_3\right|}{r^2} \\ =8.98755\times {10}^9\times \frac{\left|\left(-1.5\times {10}^{-9}\right)\left(5\times {10}^{-9}\right)\right|}{(0.2{)}^2} \\ =1.685\times {10}^{-6}\mathrm{N} \end{gathered}$$

Now, $F_2\to$force exerted by the second charge on the third charge

$$\begin{gathered} r=y_3-y_2 \\ =-0.4-0 \\ =-0.4m \end{gathered}$$

Force is repulsive in the negative y-axis

role="math" localid="1668160055220" $$\begin{gathered} F_2=K\frac{\left|q_2{q}_3\right|}{r^2} \\ =8.98755\times {10}^9\times \frac{\left|\left(3.2\times {10}^{-9}\right)\left(5\times {10}^{-9}\right)\right|}{(-0.4{)}^2} \\ =8.988\times {10}^{-7}\mathrm{N} \end{gathered}$$

Net force

$F_{1}and{F}_2$both in the negative y-axis

Therefore, the net force exerted by two charges on the third is

$$\begin{gathered} F_1+F_2=\left(1.685\times {10}^{-6}\right)+\left(8.988\times {10}^{-7}\right) \\ =2.583\times {10}^{-6}\mathrm{N} \end{gathered}$$

Therefore, the total force exerted by two charges on a third charge is $2.583\times {10}^{-6}N$in the negative y-direction.