Chapter 20, Problem 38A

To determine

Force exerted on the positive charge.

Answer to Problem 38A

  $F=98\text{N}$ in east.

Explanation of Solution

Given:

The given diagram is,

  

Negative charge, $q_1=-2\mu \text{C}$ and $q_2=-4\mu \text{C}$

Positive charge, $q_3=3\mu \text{C}$

Distance between the charges, $d_{13}=0.05\text{m west}$ and $d_{23}=0.03\text{m}\text{east}$

Formula Used:

From Coulomb’s law,

  $F=\frac{K{q}_1{q}_2}{d^2}$

Where,

  $F$ is electrostatic force

  $K$ is coulomb’s constant

  $q_1$ and $q_2$ are charges and

  $d$ is distance between the charges.

Calculation:

As, $K=9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$

So, force between $q_1\text{and}{q}_3$ is,

  $\begin{array}{c}{F}_1=\frac{9\times {10}^9\times \left(-2\times {10}^{-6}\right)\times \left(3\times {10}^{-6}\right)}{{0.05}^2}\\ =-22\text{N}\end{array}$

Thus, the force exerted on the positive charge by $q_1$ is $F_1=22\text{N}$ towards west.

Now, force between $q_3\text{and}{q}_2$ is,

  $\begin{array}{c}{F}_2=\frac{9\times {10}^9\times \left(3\times {10}^{-6}\right)\times \left(-2\times {10}^{-6}\right)}{{0.05}^2}\\ =-120\text{N}\end{array}$

Thus, the force exerted on the positive charge by $q_2$ is $F_2=120\text{N}$ towards east

As magnitude of $F_2$ is greater than $F_1$ therefore net force on $q_3$ is towards east and it is

  $F=120-22=98\text{N}\text{in east}$ .

Conclusion:

Therefore, the net electrostatic force on positive charge $q_3$ is $F=98\text{N}$ in east.