Question

Two point charges of q and 4q are kept 30 cm apart. At a distance ______ , on the straight line joining them, the intensity of electric field is zero.(A) 20 cm from 4q
(B) 7.5 cm from q(C) 15 cm from 4q
(D) 5 cm from q

Answer 1

Dear Student,

◆ Answer -

(A) 20 cm from 4q

◆ Explaination -

Let intensity of electric field be zero at distance x from charge q. Thus the point will be 30-x away from charge 4q.

For electric field intensity to be equal, electric field due to two points should be equal.

E1 = E2

k.q/x^2 = k.4q/(30-x)^2

(30-x)^2 = 4x^2

30-x = 2x

3x = 30

x = 30/3

x = 10 cm

Thus this point will be 30-10 = 20 cm from charge 4q.

Hope this was helpful...