Question

Three charges each equal to q are placed at the three corners of a square of side a. Find the electric field at the fourth corner.

Answer 1

three charges each equal to q are placed at three corners A, B and C of a square ABCD of side length a.

we have to find net electric field at D

electric field due to A on D = kq/(√2a)² = kq/2a²

electric field due to B on D = kq/a²

electric field due to C on D = kq/a²

Let kq/a² = E

then, net electric field at D = resultant of electric field due to B and C on D + electric field due to A

= √(E² + E²) + E/2

= √2 E + E/2

= (2√2 + 1)E/2

hence, electric field at fourth corner is (2√2 + 1)/2 × kq/a²

aq

Answer 2

The net electric field will be $\bold{(2 \sqrt{2}+1) \frac{k q}{2 a^{2}}}$  and it will be along the diagonal line BD.

Solution:

The electric field at the corner of the square can be calculated as follows,and diagram is shown below.

Given: Point charges are equal and are named as q

Length of square = a

Thereby the electric field at D due to the near charges A and C is given by, $E=\frac{kq}{a^{2}}$

And electric field at D due to the farthest charge B will be, $E=\frac{k q}{2 a^{2}}$  

Now as the A and C charges posing electric field are opposite and perpendicular to the electric field at D and are equal. The resultant vector of the two will be as follows.

Electric field at A and C is $E_{A C}=\sqrt{E_{A}^{2}+E_{C}^{2}}=\sqrt{2} \frac{k q}{a^{2}}$  along the diagonal BD.

Therefore, the net resultant $E_{n e t}$  from A and C is stated above and from B will be,

$\begin{array}{l}{E_{n e t}=E_{A C}+E=\sqrt{2} \frac{k q}{a^{2}}+\frac{k q}{2 a^{2}}} \\ \bold{{E_{n e t}=(2 \sqrt{2}+1) \frac{k q}{2 a^{2}}}\end{array}}$