Suggested languages for you:

Americas

Europe

Q18.7-45PE

Expert-verified
Found in: Page 665

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Using the symmetry of the arrangement, determine the direction of the force on $$q$$ in the figure below, given that $${q_a} = {q_b} = + 7.50{\rm{ }}\mu {\rm{C}}$$ and $${q_c} = {q_d} = - 7.50{\rm{ }}\mu {\rm{C}}$$. (b) Calculate the magnitude of the force on the charge $$q$$, given that the square is $$10.0{\rm{ cm}}$$ on a side and $$q = {\rm{2}}{\rm{.00 }}\mu {\rm{C}}$$.

(a) The direction of the force on $$q$$ is straight downward.

(b) The magnitude of the net force on the charge $$q$$ is $$76.37{\rm{ N}}$$.

See the step by step solution

## Net electrostatic force

Electrostatic force is a vector quantity. When two charges are separated by some distance the electrostatic force between them is given as,

$$F = \frac{{KQq}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant.

The net electrostatic force on the charge will be the vector sum of the individual charge.

## (a) Direction of the force

Due to symmetry the net force on $$q$$ will be straight down, since $${q_a}$$ and $${q_b}$$ are positive and $${q_c}$$ and $${q_d}$$ are negative with same magnitude. $${q_a}$$ and $${q_b}$$ will force the charge straight downward and $${q_c}$$ and $${q_d}$$ will pull the charge straight downward.

Hence, the direction of the force on $$q$$ is straight downward.

## (b) Magnitude of the force

The force on the charge $$q$$ is represented as,

Force on the charge $$q$$

Here, $${F_a}$$ is the repulsive force on charge $$q$$ due to $${q_a}$$, $${F_b}$$ is the repulsive force on charge $$q$$ due to $${q_b}$$, $${F_c}$$ is the attractive force on charge $$q$$ due to $${q_c}$$, and $${F_d}$$ is the attractive force on charge $$q$$ due to $${q_d}$$.

## Calculating the distance

The distance of charge $$q$$ from the charges $${q_a}$$, $${q_b}$$, $${q_c}$$ and $${q_d}$$ is,

$$r = \frac{a}{{\sqrt 2 }}$$

Here, $$a$$ is the side of the square $$\left( {a = 10.0{\rm{ }}cm} \right)$$.

Substituting all known values,

$$\begin{array}{c}r = \frac{{10.0{\rm{ cm}}}}{{\sqrt 2 }}\\ = 7.07{\rm{ cm}}\end{array}$$

## Calculating the magnitude of attractive and repulsive force

The magnitude of the repulsive force on charge $$q$$ due to $${q_a}$$ is,

$${F_a} = \frac{{Kq{q_a}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$$, $$q$$ is the magnitude of the charge at the center of the square $$\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)$$, $${q_a}$$ is the magnitude of the charge at the edge of the square $$\left( {{q_a} = 7.50{\rm{ }}\mu {\rm{C}}} \right)$$, and $$r$$ is the distance between $$q$$ and $${q_a}$$ $$\left( {r = 7.07{\rm{ cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{F_a} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 \mu C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{27 N}}\end{array}$$

The magnitude of the repulsive force on charge $$q$$ due to $${q_b}$$ is,

$${F_b} = \frac{{Kq{q_b}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$$, $$q$$ is the magnitude of the charge at the center of the square $$\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)$$, $${q_b}$$ is the magnitude of the charge at the edge of the square $$\left( {{q_b} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)$$, and $$r$$ is the distance between $$q$$ and $${q_b}$$ $$\left( {r = 7.07{\rm{ cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{F_b} &= \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ &= \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N \times }}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\& = {\rm{27 N}}\end{array}$$

The magnitude of the attractive force on charge $$q$$ due to $${q_c}$$ is,

$${F_c} = \frac{{Kq{q_c}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$$, $$q$$ is the magnitude of the charge at the center of the square $$\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)$$, $${q_c}$$ is the magnitude of the charge at the edge of the square $$\left( {{q_c} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)$$, and $$r$$ is the distance between $$q$$ and $${q_c}$$ $$\left( {r = 7.07{\rm{ cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{F_c} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\{\rm{ = 27 N}}\end{array}$$

The magnitude of the attractive force on charge $$q$$ due to $${q_d}$$ is,

$${F_d} = \frac{{Kq{q_d}}}{{{r^2}}}$$

Here, $$K$$ is the electrostatic force constant $$\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)$$, $$q$$ is the magnitude of the charge at the center of the square $$\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)$$, $${q_d}$$ is the magnitude of the charge at the edge of the square $$\left( {{q_d} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)$$, and $$r$$ is the distance between $$q$$ and $${q_d}$$ $$\left( {r = 7.07{\rm{ cm}}} \right)$$.

Substituting all known values,

$$\begin{array}{c}{F_d} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{27 N}}\end{array}$$

## Calculating the magnitude of net force

The force on the horizonal direction is,

$$\begin{array}{c}{F_x} = {F_a}\sin \left( {45^\circ } \right) - {F_b}\sin \left( {45^\circ } \right) - {F_c}\sin \left( {45^\circ } \right) + {F_d}\sin \left( {45^\circ } \right)\\ = \left( {{F_a} - {F_b} - {F_c} + {F_d}} \right) \times \sin \left( {45^\circ } \right)\end{array}$$

Substituting all known values,

$$\begin{array}{c}{F_x} = \left[ {\left( {{\rm{27 N}}} \right) - \left( {{\rm{27 N}}} \right) - \left( {{\rm{27 N}}} \right) + \left( {{\rm{27 N}}} \right)} \right] \times {\rm{sin}}\left( {{\rm{45^\circ }}} \right)\\ = 0\end{array}$$

The force on the vertical direction is,

$$\begin{array}{c}{F_y} = {F_a}\cos \left( {45^\circ } \right) + {F_b}\cos \left( {45^\circ } \right) + {F_c}\cos \left( {45^\circ } \right) + {F_d}\cos \left( {45^\circ } \right)\\ = \left( {{F_a} + {F_b} + {F_c} + {F_d}} \right) \times \cos \left( {45^\circ } \right)\end{array}$$

Substituting all known values,

$$\begin{array}{c}{F_y} = \left[ {\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right)} \right] \times {\rm{cos}}\left( {{\rm{45^\circ }}} \right)\\ = {\rm{76}}{\rm{.37 N}}\end{array}$$

The magnitude of net force on charge $$q$$ is,

$$F = \sqrt {F_x^2 + F_y^2}$$

Substituting all known values,

$$\begin{array}{c}F = \sqrt {{{\left( {\rm{0}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{\rm{76}}{\rm{.37 N}}} \right)}^{\rm{2}}}} \\ = {\rm{76}}{\rm{.37 N}}\end{array}$$

Hence, the magnitude of the net force on the charge $$q$$ is $$76.37{\rm{ N}}$$.

## Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.