Question

A test charge of \({\rm{ + 2 \mu C}}\) is placed halfway between a charge of \({\rm{ + 6 \mu C}}\) and another of \({\rm{ + 4 \mu C}}\) separated by \(10{\rm{ cm}}\). (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the \({\rm{ + 6 \mu C}}\)charge)?

Short answer

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

(b) The force is directed away from the \({\rm{ + 6 \mu C}}\) charge.

Step-by-step solution

Electrostatic force:

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.

(a) Magnitude of force on test charge:

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 \({q_2}\).

Force on test charge:

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 \(14.4{\rm{ }}N\).

(b) Direction of this force

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