Question

four point-charges are placed at the corners of a square of side 2 cm .find the magnitude and direction of the electic field at centre O of the square ,if q=0.02C

Answer 1

Answer :- zero.

Explanation :- four equal charges are placed at the corners of square of side length 2cm . We have to find out electic field at the centre of square .
We know, square is symmetrical shape and also all the given charges are equal e.g., 0.02μC . So, electric at centre of square will be zero. Because opposite nature electic field will be balanced .

For better understanding you can see in figure.
Here, it is clearly shown that electric field at O = 0

Answer 2

Seems like your question is incomplete. Answer is given according to the correct one.

Explanation:

Resultant E due to charges at A and C is:

      = |k (-2q - (-q))/r^2|    ,towards A.

      = k q/r^2     , towards A

Similarly, E due to charges at B and D is:

      = k (2q - q)/r^2   , towards B

      = k q/r^2    , towards B

Their resultant is √E²x + E²y + 2ExEycos90°

           = √(kq/r²)² + (kq/r²)²

           = (k q/r²) √2

Here, r =1/2 *diagonal of square = 1/2 *2√2 *10cm = √2 * 10^(-2) m.

Hence, resultant E is:

 = √2 k q/r^2

 = √2 * 9 * 10^(9) * 0.02  * 10^(-6)/(√2 *10^(-2))^2

 = 1.27 * 10^6  N/C

              As it is going through A and B, direction is vertically upwards.