Question

Charges q, 2 q, 3 q and 4 q are placed at the corners A, B, C and D of a square as shown in the following figure. The direction of electric field at the center of the square is along.

• A. CB
• B. AC
• C. BD
• D. AB

Answer 1

The correct option is A CB

Electric field at $P$ due to charge at $A$ is $E_A=\frac{Kq}{a^2/2}$ along $AP$

Electric field at $P$ due to charge at $B$ is $E_B=\frac{K\left(2q\right)}{a^2/2}$ along $BP$

Electric field at $P$ due to charge at $C$ is $E_C=\frac{K\left(3q\right)}{a^2/2}$ along $CP$

Electric field at $P$ due to charge at $D$ is $E_D=\frac{K\left(4q\right)}{a^2/2}$ along $DP$

Electric field, $E_A$ and $E_C$ are in opposite directions. So, their net field is

$E_{CA}=\frac{K\left(3q\right)}{a^2/2}-\frac{Kq}{a^2/2}=\frac{K\left(2q\right)}{a^2/2}$ along $CA$

Electric field $E_D$ and $E_B$ are in opposite directions, so their net field is,

$E_{DB}=\frac{K\left(4q\right)}{a^2/2}-\frac{K\left(2q\right)}{a^2/2}=K\frac{\left(2q\right)}{a^2/2}$ along $DB$

Since, $E_{CA}$ and $E_{DB}$ are equal in magnitude and they are perpendicular to each other. So, their $E$ will bisect angle $APB$.

or, we can say it is along $CB$.