Chapter 16, Problem 14P

To determine

To find: The magnitude and direction of the force on each charge.

Answer to Problem 14P

The magnitude of force on each charge is same that is $6.2\times {10}^7\text{N}$ but the direction of force is different. Force on $\text{B}$ will make ${45}^{\circ }$ with the positive $x$ axis, force on $\text{C}$ will make $-{45}^{\circ }$ and force on $\text{D}$ will make ${225}^{\circ }$ .

Explanation of Solution

Given:

A charge of $6.00\text{mC}$ is placed at each corner of a square $0.100\text{m}$ on a side.

Formula used:

Consider the figure shown below.

  

The magnitudes of the individual forces can be found using Coulomb's law. The direction of each force will be along the line connecting $q_A$ to $q_{\text{B}},q_{\text{c}}$ and $q_{\text{D}}$ . The forces at the upper left corner of the square are the vector sum of the forces due to the other three charges.

  ${\overrightarrow{F}}_A={\overrightarrow{F}}_{\text{B}}+{\overrightarrow{F}}_{\text{c}}+{\overrightarrow{F}}_{\text{D}}$

Calculation:

Let the variable $r$ represents the $0.10\text{m}$ length of a side of the square, and let the variable $q$ represents the $6.0\text{mC}$ charge at each corner.

The force on the charge at A due to the other 3 charges is,

  ${\overrightarrow{F}}_A={\overrightarrow{F}}_{\text{B}}+{\overrightarrow{F}}_{\text{c}}+{\overrightarrow{F}}_{\text{D}}$

  $=k{q}_{\wedge }\left[\frac{q_{\text{B}}(-\widehat{i})}{r_{\text{AB}}^2}+\frac{q_{\text{C}}}{r_{\text{AC}}^2}\left(\sin {45}^{\circ }(\widehat{j})+\cos {45}^{\circ }(-\widehat{i})\right)+\frac{q_{\text{D}}(\widehat{j})}{r_{\text{AD}}^2}\right]$

  $=k{q}^2\left[\left(\frac{1}{r^2}+\frac{1}{2r^2}\cos {45}^{\circ }\right)(-\widehat{i})+\left(\frac{1}{2r^2}\sin {45}^{\circ }+\frac{1}{r^2}\right)(\widehat{j})\right]$

  $=\frac{k{q}^2}{r^2}\left[\left(1+\frac{1}{2\sqrt{2}}\right)(-\widehat{i})+\left(\frac{1}{2\sqrt{2}}+1\right)(\widehat{j})\right]$

Hence,

  $\begin{array}{l}\left|{\overrightarrow{F}}_A\right|=\frac{k{q}^2}{r^2}{\left[{\left(1+\frac{1}{2\sqrt{2}}\right)}^2+{\left(\frac{1}{2\sqrt{2}}+1\right)}^2\right]}^{1/2}\\ =\frac{\sqrt{2}k{q}^2}{r^2}\left(1+\frac{1}{2\sqrt{2}}\right)\\ =\frac{k{q}^2}{r^2}\left(\sqrt{2}+\frac{1}{2}\right)\\ =\frac{\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(6\times {10}^{-3}\text{C}\right)}^2}{{(0.1\text{m})}^2}\left(\sqrt{2}+\frac{1}{2}\right)\\ =6.2\times {10}^7\text{N}\end{array}$

The direction is given as,

  $\begin{array}{cc}\theta & ={\tan }^{-1}\left(-\frac{\frac{1}{2\sqrt{2}}+1}{1+\frac{1}{2\sqrt{2}}}\right)\\ & ={135}^{\circ }\end{array}$

Here, $\theta$ is the angle made with the positive $x$ axis.

Hence, for each charge, the magnitude of net force is same that is $6.2\times {10}^7\text{N}$ but the direction of force is different.

Force on $\text{B}$ will make ${45}^{\circ }$ with the positive $x$ axis, force on $\text{C}$ will make $-{45}^{\circ }$ and force on $\text{D}$ will make ${225}^{\circ }$