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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) A charge of 6.15 mC is placed at each corner of a square 0.100 m on a side. Determine the magnitude and direction of the force on each charge.

The magnitude of the force on each charge is $$6.52 \times {10^7}\;{\rm{N}}$$. The force is directed $$45^\circ$$ diagonally away from the centre.

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

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.

## Given data

The charge at each corner is, $$q = 6.15\;{\rm{mC}} = 6.15 \times {10^{ - 3}}\;{\rm{C}}$$.

The side of the square is, $$d = 0.100\;{\rm{m}}$$.

## Determination of the magnitude of force on each charge The force on the charge at A in the x-direction is given as:

\begin{aligned}{c}{F_x} = {F_4} + {F_3}\cos 45^\circ \\{F_x} = k\frac{{{q^2}}}{{{d^2}}} + k\frac{{{q^2}}}{{{{\left( {\sqrt 2 d} \right)}^2}}} \times \frac{1}{{\sqrt 2 }}\\{F_x} = k\frac{{{q^2}}}{{{d^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\end{aligned}

Substitute the values in the above expression.

\begin{aligned}{l}{F_x} = \left( {9.0 \times {{10}^9}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {6.15 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}^2}}}{{{{\left( {0.100\;{\rm{m}}} \right)}^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\\{F_x} = 4.61 \times {10^7}\;{\rm{N}}\end{aligned}

The force on the charge at A in the y-direction is given as:

\begin{aligned}{c}{F_y} = {F_2} + {F_3}\sin 45^\circ \\{F_y} = k\frac{{{q^2}}}{{{d^2}}} + k\frac{{{q^2}}}{{{{\left( {\sqrt 2 d} \right)}^2}}} \times \frac{1}{{\sqrt 2 }}\\{F_y} = k\frac{{{q^2}}}{{{d^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\end{aligned}

Substitute the values in the above expression.

\begin{aligned}{l}{F_y} = \left( {9.0 \times {{10}^9}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {6.15 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}^2}}}{{{{\left( {0.100\;{\rm{m}}} \right)}^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\\{F_y} = 4.61 \times {10^7}\;{\rm{N}}\end{aligned}

The net force on charge A is given as:

$${F_{\rm{A}}} = \sqrt {F_x^2 + F_y^2}$$

Substitute the values in the above expression.

\begin{aligned}{c}{F_{\rm{A}}} = \sqrt {{{\left( {4.61 \times {{10}^7}\;{\rm{N}}} \right)}^2} + {{\left( {4.61 \times {{10}^7}\;{\rm{N}}} \right)}^2}} \\{F_{\rm{A}}} = 6.52 \times {10^7}\;{\rm{N}}\end{aligned}

Since all the charges are identical and they are located at the corners of a square, then from symmetry, the magnitude of the force on each charge will be the same.

Thus, the magnitude of force on each charge is $$6.52 \times {10^7}\;{\rm{N}}$$.

## Determination of the direction of force on each charge

The direction of the net force on charge A is given as:

$$\tan \alpha = \frac{{{F_{\rm{y}}}}}{{{F_{\rm{x}}}}}$$

Substitute the values in the above expression.

\begin{aligned}{c}\tan \alpha = \frac{{6.52 \times {{10}^7}\;{\rm{N}}}}{{6.52 \times {{10}^7}\;{\rm{N}}}}\\\alpha = {\tan ^{ - 1}}\left( 1 \right)\\\alpha = 45^\circ \end{aligned}

Thus, the direction of the net force on each charge is $$45^\circ$$ diagonally away from the centre. ### Want to see more solutions like these? 