Question

Four point charges Q, 2Q,3Q and 4Q are placed at the vertices of a square of side a in order . Calculate magnitude and direction of electeic field at the point of intersection of diagonals

Answer 1

Answer:

Enet= 4kQ/a^2√2 and direction is towards the side of charge Q and 2Q

Explanation: question is asking to find the electric field at the center of the square.

the formula for electric field is k q / separation ^2

here k is 9*10^9

separation = a/√2

electric field for Q = kQ/(a/√2)^2

                              = 2kQ/a^2

electric field for 2Q = 4kQ/a^2

electric field for 3Q = 6kQ/a^2

electric field for 4Q = 8kQ/a^2

now, charge Q and 3Q lies on the same diagonal

and direction of electric field of Qand 3Q are opposite to each other

so we can cancel electric field of Q nad 3Q that is remaining electric field = 4kQ/a^2 and direction is towards Q charge.

similarly, we can cancel electric field of 4Q and 2Q

so remaining charge = 4kQ/a^2 and direction is towards 2Q....

now you can see that there are two vector one is towards Q and other is towards 2Q and the angle between them is 90 degree and both vector are of equal magnitude and that is 4kQ/a^2

∴ electric field net = √ (4kQ/a^2)^2 + (4kQ/a^2)^2

                              = √2(4kQ/a^2)^2

                              = 4kQ/a^2√2 and direction is towards the side of Qand 2Q charge

hope you understood it.......................