Chapter 20.2, Problem 13PP

To determine

To Find: The net force on sphere A.

Answer to Problem 13PP

$\text{6}\text{.8}\times {\text{10}}^{-2}\text{N}$ away from it (towards the positive x -axis)

Explanation of Solution

Given:

Charge on sphere A: $q_A=+2\times {10}^{-6}\text{C}$

Sphere A is located at origin. $x_A=0$

Charge on sphere B: $q_B=-3.6\times {10}^{-6}\text{C}$

Sphere B is located at $x_B=0.60\text{m}$

Charge on sphere C: $q_C=+4.0\times {10}^{-6}\text{C}$

Sphere C is located at $x_C=0.80\text{m}$

Formula Used:

Electrostatic force between two charges can be obtained by:

$F=k\frac{q_1{q}_2}{d^2}$

Here, k is the Coulomb’s constant, $q_1$ and $q_2$ are two charges and d is the distance between them.

Calculation:

The electrostatic force between two charged spheres A and B can be calculated as follows:

  $\begin{array}{c}{F}_{AB}=k\frac{q_A{q}_B}{d_{AB}{}^2}\\ =\left(9\times {10}^9\right)\times \frac{\left(2\times {10}^{-6}\text{C}\right)\left(-3.6\times {10}^{-6}\text{C}\right)}{{0.60}^2}\\ =-1.80\times {10}^{-1}\text{N}\end{array}$

The negative sign symbolizes attractive force.

The electrostatic force between two charged spheres A and C can be calculated as follows:

  $\begin{array}{c}{F}_{AC}=k\frac{q_A{q}_C}{d_{AC}{}^2}\\ =\left(9\times {10}^9\right)\times \frac{\left(2\times {10}^{-6}\text{C}\right)\left(4.0\times {10}^{-6}\text{C}\right)}{{0.80}^2}\\ =1.12\times {10}^{-1}\text{N}\end{array}$

The net force on sphere A is:

$\begin{array}{c}{F}_A=-1.80\times {10}^{-1}\text{N}+1.12\times {10}^{-1}\text{N}\\ \text{=-6}\text{.8}\times {\text{10}}^{-2}\text{N}\end{array}$

Conclusion:

Thus, the net force on sphere A is $\text{6}\text{.8}\times {\text{10}}^{-2}\text{N}$ away from it (towards the positive x -axis).