Chapter 20, Problem 62A

To determine

The net force on sphere B.

Answer to Problem 62A

The net force on sphere B is $3.7\times {10}^2\text{ N}$ at $197°$ from the positive x- axis.

Explanation of Solution

Given:

Three charged spheres along with positionare shown in Figure below.

Formula used:

The expression for Coulomb’s law is,

$F=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}$

Here, $K$ is the Coulomb’s constant, $q_{\text{A}}\text{ and }{q}_{\text{B}}$ are the two charges, and $r$ is the distance between the charges.

Calculation:

Consider the Coulomb’s constant as $K=9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ . The force of sphere A on sphere B is,

$\begin{array}{l}{F}_1=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_1=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(+4.5\text{ μC}\times \frac{{10}^{-6}\text{ C}}{1\text{ μC}}\right)\left(-8.2\text{ μC}\times \frac{{10}^{-6}\text{ C}}{1\text{ μC}}\right)}{{\left(4.0\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)}^2}\\ F_1=-207.5\text{ N}\\ F_1=208\text{ N}\left(\text{Towards left}\right)\end{array}$

The distance between sphere B and sphere C is,

$\begin{array}{l}r=\sqrt{{\left(4.0\text{ cm}\right)}^2+{\left(3.0\text{ cm}\right)}^2}\\ r=5.0\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\\ r=0.05\text{ cm}\end{array}$

The direction of the force of sphere C on sphere B is,

$\begin{array}{l}\tan {\theta }_1=\frac{3.0\text{ cm}}{4.0\text{ cm}}\\ {\theta }_1=37°\text{below the negative }x\text{-axis}\\ {\theta }_1=180°+37°\\ {\theta }_1=217°\text{from the positive }x\text{-axis}\end{array}$

The force of sphere C on sphere B is,

$\begin{array}{l}{F}_2=K\frac{q_{\text{C}}{q}_{\text{B}}}{r^2}\\ F_2=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(+6.0\text{ μC}\times \frac{{10}^{-6}\text{ C}}{1\text{ μC}}\right)\left(-8.2\text{ μC}\times \frac{{10}^{-6}\text{ C}}{1\text{ μC}}\right)}{{\left(\text{0}\text{.05 cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)}^2}\\ F_2=-177\text{ N}\end{array}$ .

The net force along x axis is,

$\begin{array}{l}{F}_{\text{net,}\text{x}}=\left(-208\text{ N}\right)+\left(\left(-177\text{ N}\right)\cos 37°\right)\\ F_{\text{net,}\text{x}}=-349\text{ N}\\ F_{\text{net,}\text{x}}=349\text{ N}\left(\text{Towards left}\right)\end{array}$

Find the net force along y axis as follows:

$\begin{array}{l}{F}_{\text{net,}\text{y}}=\left(-177\text{ N}\right)\cos 37°\\ F_{\text{net,}\text{y}}=-106\text{ N}\\ F_{\text{net,}\text{y}}=106\text{ N}\left(\text{downwards}\right)\end{array}$

The direction of net force is,

$\begin{array}{l}\tan \theta =\frac{F_{\text{net,y}}}{F_{\text{net,x}}}\\ \tan \theta =\frac{106\text{ N}}{349\text{ N}}\\ \theta =17°\text{ below the negative }x\text{-axis}\\ \theta =180°+17°\\ \theta =197°\text{ from the positive }x\text{-axis}\end{array}$

The net force on sphere B is,

$\begin{array}{l}{F}_{\text{net}}=\sqrt{F_{\text{net,}\text{x}}^2+F_{\text{net,}\text{y}}^2}\\ F_{\text{net}}=\sqrt{{\left(-349\text{N}\right)}^2+{\left(-106\text{N}\right)}^2}\\ F_{\text{net}}=365\text{ N}\\ F_{\text{net}}=3.7\times {10}^2\text{ N}\end{array}$

Conclusion:

Thus, the net force on sphere B is $F_{\text{net}}=3.7\times {10}^2\text{ N}$ at $197°$ from the negative x -axis.