Question

Four point charges are at the corners of a square of side a as shown in Figure P15.8. Determine the magnitude and direction of the resultant electric force on q, with k_e, q, and a left in symbolic form.

Answer 1

The electric field due to a point charge at a distanceis,

Here is the Coulomb’s constant, is the charge of the point charge, and is the distance from the point charge to the field point.

The arrangement of charges is shown in the below figure.

$C:\Users\dell\Desktop\ccccx.jpg$

The electric field at point due to the charge along the horizontal axis is,

Here, is the distance of point from the charge.

The electric field at point due to charge along the horizontal direction is,

Here, is equal to the length of the diagonal of the square. That is, is the distance of the point from the charge.

The electric field at point due to charge along the vertical direction is,

Therefore, the total electric field at due to the charge is,

Here, and are the unit vectors along the horizontal and vertical axes respectively.

The electric field at due to the charge along the vertical direction is,

Here, is the distance of the point from the charge.

The net electric field at point is the sum of the electric fields due to the charges three charges in the corners of the square.

Hence, the magnitude of the net field at point is,

The direction of the net electric field is

The direction is counter-clockwise with the positive x axis.

The magnitude of the net electrostatic force is,

Substitute for.

Therefore, magnitude of the force on charge isand the direction of the net force with the horizontal is.

Answer 2

The formula for the electrostatic force is:

$$ F=\frac{k|q||Q|}{r^{2}} $$

Using the formula of electrostatic force, the force on charge \(q\) due to the other charge be:

$$ \begin{aligned} &F_{A}=k \frac{(2 q)(q)}{a^{2}}=\left(\frac{k q^{2}}{a^{2}}\right) 2 \\ &F_{C}=k \frac{(2 q)(q)}{a^{2}}=\left(\frac{k q^{2}}{a^{2}}\right) 2 \end{aligned} $$

and

$$ F_{D}=k \frac{(3 q)(q)}{(\sqrt{2} a)^{2}}=\left(\frac{k q^{2}}{a^{2}}\right)\left(\frac{3}{2}\right) $$

The magnitude of the force \(F_{A}\) and \(F_{C}\) are the same. Thus, due to symmetry, the resultant force will be along the angle bisector of the angle formed by \(F_{\mathrm{A}}\) and \(F_{\mathrm{C}} \cdot F_{\mathrm{A}}\) and \(F_{\mathrm{C}}\) are mutually perpendicular. So,

$$ \begin{aligned} F_{A C} &=\sqrt{F_{A}^{2}+F_{C}^{2}} \\ &=\sqrt{\left(\frac{k q^{2}}{a^{2}}\right)^{2}(2)^{2}+\left(\frac{k q^{2}}{a^{2}}\right)^{2}(2)^{2}} \\ &=\left(\frac{k q^{2}}{a^{2}}\right) 2 \sqrt{2} \end{aligned} $$

\(F_{\mathrm{AC}}\) and \(F_{\mathrm{D}}\) are in the same direction. Thus, the net force on charge \(q\) be:

$$ \begin{aligned} F_{n e t} &=F_{A C}+F_{D} \\ &=\left(\frac{k q^{2}}{a^{2}}\right) 2 \sqrt{2}+\left(\frac{k q^{2}}{a^{2}}\right)\left(\frac{3}{2}\right) \\ &=\left(\frac{k q^{2}}{a^{2}}\right)\left(2 \sqrt{2}+\frac{3}{2}\right) \\ &=4.33\left(\frac{k q^{2}}{a^{2}}\right) \end{aligned} $$

And the direction of the net force will be \(45^{\circ}\) in a counter-clockwise direction from the positive \(x\)-axis.