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Expert-verified Found in: Page 443 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # (II) At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q (Fig. 16–54). Determine the magnitude and direction of the force on the charge 2Q. The net force on charge 2Q is $$10.1k\frac{{{Q^2}}}{{{l^2}}}$$ at an angle $$61^\circ$$ above the positive x-axis.

See the step by step solution

## Understanding the Coulomb’s Law

Coulomb’s law states that the magnitude of the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the separation between them.

The expression for the force between two point charges is given as:

$$F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}$$ … (i)

Here, k is the Coulomb’s constant, $${Q_1},\;{Q_2}$$ are the charges and r is the separation between them.

For more than two charges, the force on any charge is the vector sum of the forces exerted due to individual charges.

## Given data

The charges at the corners are Q, 2Q, 3Q and 4Q.

The sides of the square are $$l$$.

## Determination of the magnitude of the net force on charge 2Q

The first figure below shows the direction of forces on charge 2Q due to other charges. The second figure shows the direction of the net force. The force on charge 2Q due to charge Q is,

\begin{aligned}{c}{F_1} = k\frac{{Q \times 2Q}}{{{l^2}}}\\ = 2k\frac{{{Q^2}}}{{{l^2}}}\;\;\left( {{\rm{along}}\;{\rm{x - axis}}} \right)\end{aligned}

The force on charge 2Q due to charge 3Q is,

\begin{aligned}{c}{F_3} = k\frac{{3Q \times 2Q}}{{{l^2}}}\\ = 6k\frac{{{Q^2}}}{{{l^2}}}\;\;\left( {{\rm{along}}\;{\rm{y - axis}}} \right)\end{aligned}

The length of the diagonal of the square is $$\sqrt 2 l$$.

The force on the charge 2Q due to charge 4Q is,

\begin{aligned}{c}{F_4} = k\frac{{4Q \times 2Q}}{{{{\left( {\sqrt 2 l} \right)}^2}}}\\ = k\frac{{4{Q^2}}}{{{l^2}}}\;\;\left( {{\rm{along}}\;45^\circ \;{\rm{with}}\;{\rm{x - axis}}} \right)\end{aligned}

The component of force $${F_4}$$ along the x-axis is,

\begin{aligned}{c}{F_{{\rm{4x}}}} = k\frac{{4{Q^2}}}{{{l^2}}}\cos 45^\circ \\ = k\frac{{4{Q^2}}}{{{l^2}}}\; \times \frac{1}{{\sqrt 2 }}\\ = 2\sqrt 2 \frac{{{Q^2}}}{{{l^2}}}\end{aligned}

The component of force $${F_4}$$ along the y-axis is,

\begin{aligned}{c}{F_{{\rm{4y}}}} = k\frac{{4{Q^2}}}{{{l^2}}}\sin 45^\circ \\ = k\frac{{4{Q^2}}}{{{l^2}}}\; \times \frac{1}{{\sqrt 2 }}\\ = 2\sqrt 2 \frac{{{Q^2}}}{{{l^2}}}\end{aligned}

The net force on charge 2Q along the x-axis is,

\begin{aligned}{c}{F_{\rm{x}}} = {F_{{\rm{4x}}}} + {F_1}\\ = 2\sqrt 2 \frac{{{Q^2}}}{{{l^2}}} + 2k\frac{{{Q^2}}}{{{l^2}}}\\ = 2\left( {\sqrt 2 + 1} \right)k\frac{{{Q^2}}}{{{l^2}}}\end{aligned}

The net force on charge 2Q along the y axis is,

\begin{aligned}{c}{F_{\rm{y}}} = {F_{{\rm{4y}}}} + {F_3}\\ = 2\sqrt 2 \frac{{{Q^2}}}{{{l^2}}} + 6k\frac{{{Q^2}}}{{{l^2}}}\\ = 2\left( {\sqrt 2 + 3} \right)k\frac{{{Q^2}}}{{{l^2}}}\end{aligned}

Now, the resultant force on the charge 2Q is,

\begin{aligned}{c}F = \sqrt {F_{\rm{x}}^2 + F{}_{\rm{y}}^2} \\ = \sqrt {{{\left( {2\left( {\sqrt 2 + 1} \right)k\frac{{{Q^2}}}{{{l^2}}}} \right)}^2} + {{\left( {2\left( {\sqrt 2 + 3} \right)k\frac{{{Q^2}}}{{{l^2}}}} \right)}^2}} \\ = \sqrt {{{\left( {2\left( {\sqrt 2 + 1} \right)} \right)}^2} + {{\left( {2\left( {\sqrt 2 + 3} \right)} \right)}^2}} k\frac{{{Q^2}}}{{{l^2}}}\\ = 10.1k\frac{{{Q^2}}}{{{l^2}}}\end{aligned}

Thus, the magnitude of the net force on charge 2Q is $$10.1k\frac{{{Q^2}}}{{{l^2}}}$$.

## Determination of the direction of the net force on charge 2Q

Let $$\theta$$ be the angle of the resultant force on 2Q with the x-axis.

The direction of the resultant force is calculated as:

\begin{aligned}{c}\tan \theta = \frac{{{F_{\rm{y}}}}}{{{F_{\rm{x}}}}}\\\tan \theta = \frac{{2\left( {\sqrt 2 + 3} \right)k\frac{{{Q^2}}}{{{l^2}}}}}{{2\left( {\sqrt 2 + 1} \right)k\frac{{{Q^2}}}{{{l^2}}}}}\\\tan \theta = \frac{{\sqrt 2 + 3}}{{\sqrt 2 + 1}}\\\theta = 61^\circ \end{aligned}

Thus, the resultant force on charge 2Q is directed at an angle $$61^\circ$$ above the positive x-axis. ### Want to see more solutions like these? 