The three charges $$\frac{q}{2}, q$$ and $$\frac{q}{2}$$ are placed at the corners $$A, D$$ and $$C$$ of a square of side $$a$$ as shown in figure. The magnitude of electric field ($$E$$) at the corner $$B$$ of the square, is

Charges are:

At A:

$$\frac{q}{2}$$

At D:

q

At C:

$$\frac{q}{2}$$

Side of square:

a

Find net electric field at B.

Field Charge due to A

Distance

AB=a

So

$$E_A=\frac{k(q/2)}{a^2}=\frac{kq}{2a^2}$$

Direction: along +x (to right).

Field due to Charge at C

Distance

BC=a

So

$$E_C=\frac{kq}{2a^2}$$

Direction downward (−y).

Field due to charge at D

Diagonal distance

DB= 2a

So

$$E_D=\frac{kq}{(\sqrt{2}a)^2}$$

Direction along diagonal DB, making $$45^{\circ}$$.

Components:

$$E_{Dx}=\frac{kq}{2\sqrt{\ 2}a^2}$$

$$E_{Dy}=\frac{kq}{2\sqrt{\ 2}a^2}$$
Net components

X-component:

$$E_x=\frac{kq}{2a^2}+\frac{kq}{2\sqrt{2}a^2}$$

$$=\frac{kq}{2a^2}\left(1+\frac{1}{\sqrt{2}}\right)$$

Y-component:

$$E_y=-\frac{kq}{2a^2}-\frac{kq}{2\sqrt{2}a^2}$$

$$=-\frac{kq}{2a^2}\left(1+\frac{1}{\sqrt{2}}\right)$$

So

$$|E_x|=|E_y|$$

Hence magnitude

$$E=\sqrt{\ 2}E_x$$

$$=\sqrt{2}\cdot\frac{kq}{2a^2}\left(1+\frac{1}{\sqrt{2}}\right)$$

Simplify:

$$E=\frac{kq}{2a^2}(1+\sqrt{2})$$

$$\frac{q}{4\pi\varepsilon_0 a^2}\left(\frac{1}{\sqrt{2}} + \frac{1}{2}\right)$$