A point charge of mass 0.185 kg, and net charge\( + {\bf{0}}{\bf{.340 }}\mu {\bf{C}}\)hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 16–67).

FIGURE 16–67 Problem 61.

Three forces are acting on the point charge, which is located at the end of the insulating cord.The weight of the point charge will act in the downward direction. The electric force will act in the downward direction. The tension in the insulating cord acts in the upward direction.

The given data can be listed below as:

• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The mass of the charge is\(m = 0.185{\rm{ kg}}\).
• The value of the point charge is\(Q = + 0.340{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6{\rm{ }}}}{\rm{C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = 3.4 \times {10^{ - 7}}{\rm{ C}}\).
• The tension force in the insulating cord is\({F_{\rm{T}}} = 5.18{\rm{ N}}\).
• The Coulomb’s constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

The weight of the charge can be expressed as:

\(W = mg\)

Substitute the values in the above equation.

\(\begin{aligned}{c}W &= 0.185{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ &= 1.82{\rm{ N}}\end{aligned}\)

The tension in the insulating cord is 2 N. The weight of the charge is 1.82 N. The tension in the cord is more than the weight of the charge.

So, the electric force should be in the direction opposite to the tension force. The electric field should be in the downward direction.

The free body diagram can be shown as:

Here,\(Q\)is the charge, mg is the weight of the charge, \({F_{\rm{E}}}\) is the electric force, and \({F_{\rm{T}}}\) is the tension force in the cord.

At equilibrium, resolve the forces along the vertical axis. The net force acting in the vertical can be equated to zero.

The force acting along the vertical direction can be expressed as:

\(\begin{aligned}{c}\sum {{F_y}} &= 0\\{F_{\rm{T}}} - mg - {F_{\rm{E}}} &= 0\end{aligned}\)

\({F_{\rm{E}}} = {F_{\rm{T}}} - mg\) … (i)

Substitute the values in the above equation.

\(\begin{aligned}{c}{F_{\rm{E}}} &= 5.18{\rm{ N}} - 0.185{\rm{ kg}} \times 9.81{\rm{ m/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ &= 5.18\;{\rm{N}} - 1.82\;{\rm{N}}\\ &= 3.36\;{\rm{N}}\end{aligned}\)

The electric force on the point charge can also be expressed as:

\(\begin{aligned}{c}{F_{\rm{E}}} &= QE\\E &= \frac{{{F_{\rm{E}}}}}{Q}\end{aligned}\) … (ii)

Substitute the values in the above equation.

\(\begin{aligned}{c}E &= \frac{{3.36{\rm{ N}}}}{{3.4 \times {{10}^{ - 7}}{\rm{ C}}}}\\ &= 9.88 \times {10^6}{\rm{ N/C}}\end{aligned}\)

Thus, the magnitude of the electric field is \(9.88 \times {10^6}{\rm{ N/C}}\) and the direction of the electric field is downward.