Question

Two-point charges are placed on the x-axis as follows: Charge${\mathit{q}}_{\mathbf{1}}$= +4.00 nC is located at x = 0.200 m, and charge${\mathit{q}}_{\mathbf{2}}$= +5.00 nC is at x = -0.300 m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge${\mathit{q}}_{\mathbf{3}}$= -6.00 nC that is placed at the origin?

Short answer

The magnitude of the net force is $F_{net}=2.4\mu N$and the direction of the net force is positive x-direction.

Step-by-step solution

Coulomb’s 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

$$\begin{gathered} r=x_3-x_1 \\ =0-0.2 \\ =-0.2m \end{gathered}$$

Here $q_1$is a positive and $q_3$is a negative

Therefore force $F_1$is towards the positive axis

$$\begin{gathered} F_1=K\frac{\left|q_1{q}_2\right|}{r^2} \\ =\frac{8.9875\times {10}^9\times \left|\left(4\times {10}^9\right)\left(-6\times {10}^9\right)\right|}{(0.2{)}^2} \\ =5.393\times {10}^{-6}\mathrm{N} \end{gathered}$$

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

$$\begin{gathered} r=x_3-x_2 \\ =0-(-0.3) \\ =0.3m \end{gathered}$$

Here $q_2$is the positive and $q_3$is negative.

Therefore force $F_2$is towards the negative axis.

Magnitude and direction of the net force

$F_1\to$towards positive x-axis

$F_2\to$towards the negative x-axis.

$$\begin{gathered} F_1+F_2=5.393\times {10}^{-6}-2.995\times {10}^{-6} \\ =2.398\times {10}^{-6} \\ \approx 2.4\mu C \end{gathered}$$

Hence, The magnitude of the net force is $F_{net}=2.4\mu N$and the direction of the net force is positive x-direction.