Chapter 20, Problem 3STP

To determine

To Choose: The correct option.

Answer to Problem 3STP

Option A: $78\text{N}$ towards charge A

Explanation of Solution

Given:

The figure depicting the charges and their respective distances.

The charge A, $q_A=+8.5\times {10}^{-6}\text{}\text{C}$

The charge B, $q_B=+3.1\times {10}^{-6}\text{}\text{C}$

The charge C, $q_C=+6.4\times {10}^{-6}\text{}\text{C}$

The distance between charge A and B, $r_{AB}=4.2\text{cm = 0}\text{.042 m}$

The distance between charge B and C, $r_{BC}=2.9\text{cm = 0}\text{.029 m}$

Formula used:

The electrostatic force between two charges, $q_1$ and $q_2$ , $r$ distance apart is given by,

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Where ${\epsilon }_0=\text{ permittivity of free space}=8.85\times {10}^{-12}\text{F/m}$

Calculation:

Let the direction towards right be positive while that towards left be negative.

Since all the charges are positive, thus they all repel each other. Therefore, the net force on charge B will be,

  $\begin{array}{c}{F}_B=F_{AB}-F_{BC}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{q_A{q}_B}{r_{AB}{}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q_B{q}_C}{r_{BC}{}^2}\\ =\frac{q_B}{4\pi {\epsilon }_0}\left(\frac{q_A}{r_{AB}{}^2}-\frac{q_C}{r_{BC}{}^2}\right)\end{array}$

Using given values in above,

  $\begin{array}{c}{F}_B=\frac{3.1\times {10}^{-6}}{4\pi \left(8.85\times {10}^{-12}\right)}\left(\frac{8.5\times {10}^{-6}}{{\left(0.042\right)}^2}-\frac{6.4\times {10}^{-6}}{{\left(0.029\right)}^2}\right)\\ =0.027\left(4818.59-7609.98\right)\\ \approx -78\text{N}\end{array}$

Negative sign denotes towards left, that is towards charge A.

Conclusion:

Hence, the net force on the charge B is $78\text{N}$ towards charge A.