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Found in: Page 665

### College Physics (Urone)

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

# What is the force on the charge located at $$x = 8.00{\rm{ }}cm$$ in Figure 18.52(a) given that $$q = 1.00{\rm{ }}\mu C$$?Figure 18.52 (a) Point charges located at ${\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}$ along the x-axis. (b) Point charges located at ${\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}$ along the x-axis

The force on the charge located at $$x = 8.00{\rm{ cm}}$$ is $${\rm{12}}{\rm{.8 }}N$$.

See the step by step solution

## Step 1: Electrostatic force

The electrostatic force is a vector quantity. When a test charge is placed in a system of charges, each charge will exert electrostatic force on it. The resultant force on the test charge can be obtained by the vector sum of all the forces acting on it.

## Step 2: Force Diagram

The electrostatic force between two charges $${q_1}$$ and $${q_2}$$ separated by a distance r is,

$$F = \frac{{K{q_1}{q_2}}}{{{r^2}}}$$

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

The force acting on the charge located at $$x = 8.00{\rm{ cm}}$$ is represented as,

Force acting on the charge located at $$x = 8.00{\rm{ cm}}$$

Here, $${F_3}$$ is the force of attraction due to charge at $$x = 3.00{\rm{ cm}}$$ and $${F_{11}}$$ is the force of attraction due to charge at $$x = 11.00{\rm{ cm}}$$.

## Step 3: Net force

The force of attraction between the charge located at $$x = 3.00{\rm{ cm}}$$ and $$x = 8.00{\rm{ cm}}$$ is,

${F_{11}} = \frac{{K\left( q \right)\left( {2q} \right)}}{{{{\left( {{r_3} - {r_8}} \right)}^2}}}$

Substitute $$9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$$ for K, $$1.00{\rm{ }}\mu C$$ for $$q$$, $$3.00{\rm{ cm}}$$ for ${r_3}$ and $$8.00{\rm{ cm}}$$ for ${r_8}$,

$$\begin{array}{c}{{\rm{F}}_{\rm{3}}}{\rm{ = }}\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.00 \mu C}}} \right){\rm{ \times }}\left( {{\rm{2 \times 1}}{\rm{.00 \mu C}}} \right)}}{{{{\left[ {\left( {{\rm{3}}{\rm{.00 cm}}} \right){\rm{ - }}\left( {{\rm{8}}{\rm{.00 cm}}} \right)} \right]}^{\rm{2}}}}}\\{\rm{ = }}\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.00 \mu C}}} \right){\rm{ \times }}\left( {{\rm{2 \times 1}}{\rm{.00 \mu C}}} \right)}}{{{{\left[ {{\rm{ - }}\left( {{\rm{5}}{\rm{.00 cm}}} \right)} \right]}^{\rm{2}}}}}\\{\rm{ = }}\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){\rm{ \times }}\left( {{\rm{1}}{\rm{.00 \mu C}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right){\rm{ \times }}\left( {{\rm{2 \times 1}}{\rm{.00 \mu C}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left[ {{\rm{ - }}\left( {{\rm{5}}{\rm{.00 cm}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\{\rm{ = 7}}{\rm{.2 N}}\end{array}$$

The force of attraction between the charge located at $$x = 8.00{\rm{ cm}}$$ and $$x = 11.00{\rm{ cm}}$$ is,

${F_{11}} = \frac{{K\left( {2q} \right)\left( q \right)}}{{{{\left( {{r_8} - {r_{11}}} \right)}^2}}}$

Substitute $$9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$$ for $$K$$, $$1.00{\rm{ }}\mu C$$ for $$q$$, and $$8.00{\rm{ cm}}$$ for ${r_8}$, and $$11.00{\rm{ cm}}$$ for ${r_{11}}$

$$\begin{array}{c}{F_{11}} = \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \cdot {m^2}/{C^2}} \right) \times \left( {2 \times 1.00{\rm{ }}\mu C} \right) \times \left( {1.00{\rm{ }}\mu C} \right)}}{{{{\left[ {\left( {8.00{\rm{ }}cm} \right) - \left( {11.00{\rm{ }}cm} \right)} \right]}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \cdot {m^2}/{C^2}} \right) \times \left( {2 \times 1.00{\rm{ }}\mu C} \right) \times \left( {1.00{\rm{ }}\mu C} \right)}}{{{{\left[ { - \left( {3.00{\rm{ }}cm} \right)} \right]}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \cdot {m^2}/{C^2}} \right) \times \left( {2 \times 1.00{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right) \times \left( {1.00{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right)}}{{{{\left[ { - \left( {3.00{\rm{ }}cm} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}\\ = 20{\rm{ }}N\end{array}$$

The net force acting on the charge located at $$x = 8.00{\rm{ cm}}$$ is,

$$F = \left| {{F_3} - {F_{11}}} \right|$$

Substitute $${\rm{7}}{\rm{.2 }}N$$ for $${{\rm{F}}_{\rm{3}}}$$ and $$20{\rm{ }}N$$ for $${F_{11}}$$,

$$\begin{array}{c}F = \left| {\left( {7.2{\rm{ N}}} \right) - \left( {20{\rm{ N}}} \right)} \right|\\ = 12.8{\rm{ N}}\end{array}$$

Hence, force on the charge located at $$x = 8.00{\rm{ cm}}$$ is $${\rm{12}}{\rm{.8 }}N$$.