Question

Two point charges +2 C and +8 C are placed 12 cm apart in air. Find the position on the line joining these charges where the resultant electric field intensity is Zero

Answer 1

Answer:

12cm away from the charge 2 C

Explanation:

let the point on which net electric field is zero be x distance in right along the line from +2C

now we know that electric Field is Kq1q2/r^2

now E1+E2=0

K(2)/x^2=-K(8)/(12-x)^2

(1/x)^2=-(2/12-x)^2

taking under root

1/x=-2/12-x

cross multiply

-2x=12-x

x=-12

ie the point is 12 cm away from the 2 coulomb implies left of the charge 2 C