Question

In Example 21.3, calculate the net force on charge${\mathit{q}}_{\mathbf{1}}$

Short answer

The net force on the charge $q_1$is $18.0\times {10}^{-5}N$.

Step-by-step solution

Step 1:

As $q_2$exerting an attractive force on charge $q_1$

And $q_3$exerting a repulsive force in charge $q_1$.

So, combining the two forces to find the net force exerted on $q_1$.

Step 2:

Force by $q_2$on $q_1$is

$F_2=k\frac{\left|q_1{q}_2\right|}{r^2}$

Here, K is $9.0\times {10}^{9}N·m^2/C^2$, r is the distance between $q_1$and $q_2$, $r=2cm$

$q_1=1.0nC$and $q_2=3.0nC$

Putting all these values

$$\begin{gathered} F_2=\left(9.0\times {10}^9\right)\left(\frac{\left(-3\times {10}^{-9}\right)\left(1\times {10}^{-9}\right)}{{\left(0.02\right)}^2}\right) \\ =6.75\times {10}^{-5}N \end{gathered}$$

Step 3:

Force by $q_3$on $q_1$is

$F_3=K\frac{\left|q_3{q}_1\right|}{r^2}$

Here, $q_1=1.0nC$and$q_3=5.0nC$

Putting all these values

$$\begin{gathered} F_3=\left(9.0\times {10}^9\right)\left(\frac{\left(5\times {10}^{-9}\right)\left(1\times {10}^{-9}\right)}{{\left(0.02\right)}^2}\right) \\ =11.25\times {10}^{-5}N \end{gathered}$$

So, the net force is

$$\begin{gathered} F_{net}=F_2+F_3 \\ =6.75\times {10}^{-9}+11.25\times {10}^{-9} \\ =18.0\times {10}^{-5}N \end{gathered}$$

Hence, the Net force on the charge$q_1$ is $18.0\times {10}^{-5}N$ .