Chapter 16, Problem 12P

To determine

To Find: The net force on each charge due to the other two.

Explanation of Solution

Given:

Charge on particle 1, $q_1=+75\text{μC}$

Charge on particle 2, $q_2=+48\text{μC}$

Charge on particle 3, $q_3=-85\text{μC}$

Distance between particle 1 and 2, $d_{12}=0.35\text{m}$

Distance between particle 1 and 3, $d_{13}=0.70\text{m}$

Distance between particle 2 and 3, $d_{23}=0.35\text{m}$

Formula used:

Electric force can be obtained by:

  $F=\frac{k{q}_1{q}_2}{r^2}$

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

Calculation:

Force on particle 1 due to other two:

  $\begin{array}{l}{F}_{12}+F_{13}=\frac{k{q}_1{q}_2}{d_{12}{}^2}+\frac{k{q}_1{q}_3}{d_{13}{}^2}\\ F_{12}+F_{13}=9\times {10}^9\times 75\times {10}^{-6}\left(\frac{48\times {10}^{-6}}{{0.35}^2}+\frac{-85\times {10}^{-6}}{{0.70}^2}\right)\\ F_{12}+F_{13}=675\times {10}^{-3}\left(391.84-173.47\right)\\ F_{12}+F_{13}=147.4\text{N}\end{array}$

Force on particle 2 due to other two:

  $\begin{array}{l}{F}_{21}+F_{23}=\frac{k{q}_1{q}_2}{d_{12}{}^2}+\frac{k{q}_2{q}_3}{d_{23}{}^2}\\ F_{21}+F_{23}=9\times {10}^9\times 48\times {10}^{-6}\left(\frac{75\times {10}^{-6}}{{0.35}^2}+\frac{-85\times {10}^{-6}}{{0.35}^2}\right)\\ F_{21}+F_{23}=432\times {10}^{-3}\left(612.24-693.88\right)\\ F_{21}+F_{23}=-35.27\text{N}\end{array}$

Force on particle 3 due to other two:

  $\begin{array}{l}{F}_{13}+F_{23}=\frac{k{q}_1{q}_3}{d_{13}{}^2}+\frac{k{q}_2{q}_3}{d_{23}{}^2}\\ F_{13}+F_{23}=9\times {10}^9\times \left(-85\times {10}^{-6}\right)\left(\frac{75\times {10}^{-6}}{{0.70}^2}+\frac{48\times {10}^{-6}}{{0.35}^2}\right)\\ F_{13}+F_{23}=-765\times {10}^{-3}\left(153.06+391.84\right)\\ F_{13}+F_{23}=-416.85\text{N}\end{array}$