Chapter 20.2, Problem 14PP

To determine

The net force on sphere $B$ .

Answer to Problem 14PP

The net force on B is 2 N along the right-hand side of the y -axis.

Explanation of Solution

Given:

Charge on shphere A is $2.0\times {10}^{-6}\text{C}$ .

Charge on the sphere B is $3.6\times {10}^{-6}\text{C}$ .

Separation between the charges is 0.60 m.

Formula used:

The magnitude of the force between the two charges is given by

  $F=K\frac{q_{\text{A}}{q}_{\text{B}}}{r_{\text{AB}}^2}$  ......(1)

Calculation:

The problem can be analyzed as follows:

  

Force of a sphere A on sphere B , is

  $F_{\text{A}\text{}\text{on}\text{B}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r_{\text{AB}}^2}$  ......(2)

Substitute the values of $q_{\text{A}},q_{\text{B}}$ and $r$ in equation (2)

  $\begin{array}{l}{F}_{\text{A}\text{}\text{on}\text{B}}=\left(9.0\times {10}^9\right)\times \frac{\left(2.0\times {10}^{-6}\right)\times \left(3.6\times {10}^{-6}\right)}{{\left(0.60\text{m}\right)}^2}\\ F_{\text{A}\text{}\text{on}\text{B}}=18\text{N}\end{array}$

The force of sphere C on sphere B , is

  $F_{\text{C}\text{}\text{on}\text{B}}=K\frac{q_{\text{B}}{q}_{\text{C}}}{r_{\text{BC}}^2}$  ......(3)

Substitute the values and force is calculated as

  $\begin{array}{l}{F}_{\text{C}\text{}\text{on}\text{B}}=\left(9.0\times {10}^9\right)\times \frac{\left(-4.0\times {10}^{-6}\right)\times \left(3.6\times {10}^{-6}\right)}{{\left(0.80\text{m}\right)}^2}\\ F_{\text{C}\text{}\text{on}\text{B}}=20\text{N}\end{array}$

The net force on a sphere A is given by the difference between the force of B and force of C .

  $\begin{array}{c}{F}_{\text{net}}=F_{\text{C}\text{on}\text{B}}-F_{\text{A}\text{}\text{on}\text{B}}\\ =\left(20-18\right)\text{N}\\ F_{\text{net}}\text{=2}\text{N}\end{array}$

The direction of force will be same as the direction of greater force that is, the force of C on B.

Conclusion:

Hence, the net force on B is 2 N along the right-hand side of the y -axis.