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Q. 20, 21, 23, 24, 26, 29​

Answer 1

Hey mate here is your answer 》》》》

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

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