Three charges, each equal to $q$, are placed at the three corners of a square of side $a$. The magni

Question

Three charges, each equal to $q$, are placed at the three corners of a square of side $a$. The magnitude of electric field at the fourth corner is

Professor PC Thomas Classes · Answer

Let the charges be placed at the corners $A, B$ and $C$. We shall calculate the electric field at the fourth corner $D$. The field $E_1$ due to the charge at $A$ will have the magnitude $\frac{q}{4 \pi \varepsilon_0 a^2}$ and will be along $A D$. The field $E_2$ due to the charge at $C$ will have the same magnitude and will be along $C D$.
The field $E_3$ due to the charge at $B$ will have the magnitude $\frac{q}{4 \pi \varepsilon_0(\sqrt{ } 2 a)^2}$ and will be along $B D$. As $E_1$ and $E_2$ are equal in magnitude, their resultant will be along the bisector of the angle between $E_1, E_2$ and hence along $E_3$. The magnitude of this resultant is $\sqrt{E_1{ }^2+E_2{ }^2}$ as the angle between $E_1$ and $E_2$ is $\pi / 2$. The resultant electric field at $D$ is, therefore, along $E_3$ and has magnitude
$$
\begin{aligned}
& \sqrt{E_1^2+E_2^2}+E_3 \
= & \sqrt{\left(\frac{q}{4 \pi \varepsilon_0 a^2}ight)^2+\left(\frac{q}{4 \pi \varepsilon_0 a^2}ight)^2}+\frac{q}{4 \pi \varepsilon_0(\sqrt{ } 2 a)^2} \
= & \frac{q}{4 \pi \varepsilon_0}\left[\frac{\sqrt{ } 2}{a^2}+\frac{1}{2 a^2}ight]=(2 \sqrt{ } 2+1) \frac{q}{8 \pi \varepsilon_0 a^2} .
\end{aligned}
$$