Chapter 16, Problem 16P

(a)

To determine

To Find: The net force on 2Q charge.

Answer to Problem 16P

  ${\overrightarrow{F}}_{2Q}=\frac{2k{Q}^2}{l^2}\left[\left(1+\sqrt{2}\right)\widehat{i}+\left(3+\sqrt{2}\right)\widehat{j}\right]$

Explanation of Solution

Given:

The below figure represents the schematic arrangement of charges $(Q,2Q,3Q,4Q)$ which are placed at the corners of a square of side $l$

Formula Used:

Coulomb’s force:

  $F=k\frac{qq\text{'}}{r^2}$

Here, q and q’ are the charges, r is the distance between them and k is the Coulomb’s constant.

Calculation:

The direction of the force acting at the charge $2Q$ is also represented in the figure:

  

Here, Q is the charge.

The expression for the force on the charge $3Q$ due to the other 3 charges is,

  $\begin{array}{l}{\overrightarrow{F}}_{2Q}={\overrightarrow{F}}_Q+{\overrightarrow{F}}_{3Q}+{\overrightarrow{F}}_{4Q}\\ {\overrightarrow{F}}_Q=k\frac{Q\left(2Q\right)}{l^2}\widehat{i}\\ {\overrightarrow{F}}_{3Q}=k\frac{3Q\left(2Q\right)}{l^2}\widehat{j}\\ {\overrightarrow{F}}_{4Q}=k\frac{4Q\left(2Q\right)}{2l^2}\cos 45°\widehat{i}+k\frac{4Q\left(2Q\right)}{2l^2}\sin 45°\widehat{j}\\ {\overrightarrow{F}}_{2Q}=\left(k\frac{Q\left(2Q\right)}{l^2}+k\frac{4Q\left(2Q\right)}{2l^2}\cos 45°\right)\widehat{i}+\left(k\frac{3Q\left(2Q\right)}{l^2}+k\frac{4Q\left(2Q\right)}{2l^2}\sin 45°\right)\widehat{j}\\ {\overrightarrow{F}}_{2Q}=\frac{2k{Q}^2}{l^2}\left[\left(1+\sqrt{2}\right)\widehat{i}+\left(3+\sqrt{2}\right)\widehat{j}\right]\end{array}$

(b)

To determine

To Find: The net force on $\text{3Q}$ due to the other 3 charges.

Answer to Problem 16P

  ${\overrightarrow{F}}_{3Q}=\frac{3k{Q}^2}{l^2}\left(4+\frac{1}{2\sqrt{2}}\right)\widehat{i}-\frac{3k{Q}^2}{l^2}\left(2+\frac{1}{2\sqrt{2}}\right)\widehat{j}$

Explanation of Solution

Given:

The below figure represents the schematic arrangement of charges $(Q,2Q,3Q,4Q)$ which are placed at the corners of a square of side $l$ .

Formula Used:

Coulomb’s force:

  $F=k\frac{qq\text{'}}{r^2}$

Here, q and q’ are the charges, r is the distance between them and k is the Coulomb’s constant.

Calculation:

The direction of the force acting at the charge $3Q$ is also represented in the figure:

  

Here, Q is the charge.

The expression for the force on the charge $3Q$ due to the other 3 charges is,

  $\begin{array}{l}{\overrightarrow{F}}_{3Q}={\overrightarrow{F}}_Q+{\overrightarrow{F}}_{2Q}+{\overrightarrow{F}}_{4Q}\\ {\overrightarrow{F}}_Q=k\frac{Q\left(3Q\right)}{l^2}\cos 45°\widehat{i}+k\frac{Q\left(3Q\right)}{l^2}\sin 45°\left(-\widehat{j}\right)\\ {\overrightarrow{F}}_{2Q}=k\frac{2Q\left(3Q\right)}{l^2}\left(-\widehat{j}\right)\\ {\overrightarrow{F}}_{4Q}=k\frac{4Q\left(3Q\right)}{l^2}\widehat{i}\\ {\overrightarrow{F}}_{3Q}=\left(k\frac{4Q\left(3Q\right)}{l^2}+k\frac{Q\left(3Q\right)}{2l^2}\cos 45°\right)\widehat{i}-\left(k\frac{2Q\left(3Q\right)}{l^2}+k\frac{Q\left(3Q\right)}{2l^2}\sin 45°\right)\widehat{j}\\ {\overrightarrow{F}}_{3Q}=\frac{3k{Q}^2}{l^2}\left(4+\frac{1}{2\sqrt{2}}\right)\widehat{i}-\frac{3k{Q}^2}{l^2}\left(2+\frac{1}{2\sqrt{2}}\right)\widehat{j}\end{array}$