Question

Four particles carrying charges + 2q, − 2q, + 4q and − 4q (with q = 1.0 nC) are kept at the vertices of a square of side 6.0 cm. Determine the net electric field due to these charged particles at the centre of the square. What is the electrostatic force on a particle carrying positive charge of 1.0 nC placed at the centre of the square?

Answer 1

Since the electric field follows the superposition principle, we can only calculate magnitudes of its components provided by each charge and add these components as vectors. The distance from each charge to the center equals half of the diagonal of square:

$$r=a\sqrt{2}/2.$$

Let's begin.

$$E_1=k\frac{q_1}{r^2}=2k\frac{q_1}{a^2},\\ \space\\ E_2=2k\frac{q_2}{a^2},\\ \space\\ E_3=2k\frac{q_3}{a^2},\\ \space\\ E_4=2k\frac{q_4}{a^2}.\\ \space\\$$

Assume that positive x-direction is to the right and positive y upward. Then, assuming that charge 1 is in the top left corner, 2 in top right, 3 in lower right, 4 lower left corner:

(all charges are shown with their field lines closest to the charges)

$$E_\text{net.x}=\text{cos}45^\circ\cdot\frac{2k}{a^2}(q_1-q_2-q_3+q_4)=\\ \space\\ =\text{cos}45^\circ\cdot\frac{2k}{a^2}[2q-(-2q)-4q+(-4q)]=\\ \space\\ =\text{cos}45^\circ\cdot\frac{2k}{a^2}\cdot(-4q)=-14142.136\text{ V/m}. \space\\ \space\\ E_\text{net.y}=\text{cos}45^\circ\cdot\frac{2k}{a^2}(-q_1-q_2+q_3+q_4)=\\ \space\\ =\text{cos}45^\circ\cdot\frac{2k}{a^2}[-2q-(-2q)+4q+(-4q)]=0.$$

What a wonderful y-component! Thus the net field vector fully lies along x-coordinate and directed to negative x.

The electrostatic force will be

$$F=Eq=-14142.136\cdot1\cdot10^{-9}=-1.414\cdot10^{-5}\text{ N}.$$

It is directed likewise.