Question

Electric field intensity at the center of a square is _____ if +20 esu charges are placed at each corner of the square having side-length as 10 cm.

(a) 0

(b) 0.4 dyne/esu

(c) 2 dyne/esu

(d) 1.6 dyne/esu

The question was asked in unit test.

The question is from Electric Field topic in portion Charges and Fields of Physics – Class 12

Answer 1

The correct answer is (a) 0

To explain I would say: Distance of center from each corner point of the square is = \(\frac {10\sqrt2}{2}\) = 5√2. Therefore field intensity at the center due to a single charge is = \(\frac {20}{(5\sqrt2)^2}\) dyne/esu. But the fields due to the four charges are equal and are at perpendicular to each other. So the fields balance each other and the net electric field at the center will be equal to zero.