91Ó°ÊÓ

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.

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 \(x = 11.00{\rm{ cm}}\).

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 \({\rm{12}}{\rm{.8 }}N\).