91Ó°ÊÓ

Every charged object in the universe attracted other charged body in the universe with a force known as electrostatic force.

The electrostatic force between two point charges is given as,

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\) ….. (1)

Here, \(F\) is the electrostatic force, \(K\) is the electrostatic force constant, \({q_1}\) is the point charge, \({q_2}\) is the point charge, and \(r\) is the separation between the charges.

Forces on test charge is represented as,

Here, \(q\) is the test charge, \(r\) is the separation between charges \({q_1}\) and \({q_2}\), \({F_1}\) is the force on test charge \(q\) due to charge \({q_1}\), and \({F_2}\) is the force on test charge \(q\) due to \({q_2}\).

The force on test charge \(q\) due to charge \({q_1}\) is,

\({F_1} = \frac{{Kq{q_1}}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\) ….. (2)

Here, \(F\) is the electrostatic force, \(K\) is the electrostatic force constant, \({q_1}\) is the charge on the first object, \(q\) is the test charge, and \(r\) is the separation between the charges \({q_1}\) and \({q_2}\).

Consider the known data as below.

The separation between the charges, \(r = 10{\rm{ c}}m\)

The electrostatic force constant, \(K={9\times10^9}\;{\rm{N\times{m^2}/C}}\) The charge on the first body, \({q_1} = 6{\rm{ }}\mu C\)

The test charge, \(q = 2{\rm{ }}\mu C\)

Substituting all known values into equation (2).

\(\begin{aligned} {F_1} &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {2{\rm{ }}\mu C} \right) \times \left( {6{\rm{ }}\mu C} \right)}}{{{{\left( {\frac{{10{\rm{ }}cm}}{2}} \right)}^2}}}\\ &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {2{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right) \times \left( {6{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right)}}{{{{\left( {\left( {\frac{{10{\rm{ }}cm}}{2}} \right) \times \left( {\frac{{1{\rm{ }}m}}{{100{\rm{ }}cm}}} \right)} \right)}^2}}}\\ &= 43.2{\rm{ }}N\end{aligned}\)

The force on test charge \(q\) due to charge \({q_2}\) is,

\({F_2} = \frac{{Kq{q_2}}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\) ….. (3)

Here, \({q_2}\) is the charge on the second object, \(q\) is the test charge, and \(r\) is the separation between the charges \({q_1}\) and \({q_2}\).

Consider the known data as below.

The separation between the charges, \(r = 10{\rm{ c}}m\)

The electrostatic force constant, \(K={9\times10^9}\;{\rm{N\times{m^2}/C}}\)

The charge on the second body, \({q_2} = 4{\rm{ }}\mu C\)

The test charge, \(q = 2{\rm{ }}\mu C\)

Substituting all known values into equation (3).

\(\begin{aligned} {F_1} &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {2{\rm{ }}\mu C} \right) \times \left( {4{\rm{ }}\mu C} \right)}}{{{{\left( {\frac{{10{\rm{ }}cm}}{2}} \right)}^2}}}\\ &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {2{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right) \times \left( {4{\rm{ }}\mu C} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}C}}{{1{\rm{ }}\mu C}}} \right)}}{{{{\left( {\left( {\frac{{10{\rm{ }}cm}}{2}} \right) \times \left( {\frac{{1{\rm{ }}m}}{{100{\rm{ }}cm}}} \right)} \right)}^2}}}\\ &= 28.8{\rm{ }}N\end{aligned}\)

The net force on test charge is,

\(F = \left| {{F_1} - {F_2}} \right|\)

Substituting all known values,

\(\begin{aligned} F &= \left| {\left( {43.2{\rm{ }}N} \right) - \left( {28.8{\rm{ }}N} \right)} \right|\\ &= 14.4{\rm{ }}N\end{aligned}\)

Hence, the magnitude of the force on the test charge is \(14.4{\rm{ }}N\).

Since \({F_1} > {F_2}\), hence the force is directed away from the \( + 6{\rm{ }}\mu C\) charge.