Question

(Figure 1) shows four charges at the corners of a square of side L.

What magnitude and sign of charge Q will make the force on charge q zero?

Q =

Answer 1

The following figure shows the schematic diagram of the arrangement of charges at the vertices of the square of side \(L\)

In the above figure \(\vec{F}_{1}\) is the force on charge \(q\) due to \(-10 \mathrm{nC}\) charge, \(\vec{F}_{2}\) is the force on

charge \(q\) due to the charge \(-10 \mathrm{nC}, \vec{F}_{3}\) is the force on charge \(q\) due to the charge \(Q,\) and \(\vec{F}_{12}\) is the net force of \(\vec{F}_{1}\) and \(\vec{F}_{2}\)

 

As the charge \(-10 \mathrm{nC}\) pulls the charge \(q\) in the directions shown in the figure. The equilibrium is possible only when the sign of charge \(q\) is positive.

since \(\vec{F}_{1}\) and \(\vec{F}_{2}\) are mutually perpendicular,

\(F_{12}=\sqrt{F_{1}^{2}+F_{2}^{2}}\)

The magnitude of force between the charges \(-10 \mathrm{nC}\) and \(q\) is, $$ \begin{array}{l} F_{1}=k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}} \\ F_{2}=k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}} \\ F_{12}=\sqrt{\left(k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}}\right)^{2}+\left(k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}}\right)^{2}} \\ F_{12}=\sqrt{2}\left(k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}}\right) \end{array} $$ The magnitude of force between the charges \(Q\) and \(q\) is, $$ \begin{array}{l} F_{3}=k \frac{Q q}{(\sqrt{2} L)^{2}} \\ F_{3}=k \frac{Q q}{2 L^{2}} \end{array} $$ From the figure, the forces \(F_{3}\) and \(F_{12}\) are in opposite direction and hence, $$ \begin{aligned} k \frac{Q q}{2 L^{2}} &=\sqrt{2}\left(k \frac{(-10 \mathrm{n} \mathrm{C}) q}{L^{2}}\right) \\ \frac{Q}{2} &=\sqrt{2}(-10 \mathrm{n} \mathrm{C}) \\ Q &=28.3 \mathrm{n} \mathrm{C} \end{aligned} $$ Thus, the magnitude of charge is \(28.3 \mathrm{nC}\).