Question

Two types charge qp=+2uc and qq=-2uc are placed at a distance of 10cm in air/vacuum
(1) Calculate the electric field intensity along the center of line joining the two changes.
(2)If a test qr=-1.2×10*-9c is placed at center of the PQ write the value of force acting on test charge qR and it's direction​

Answer 1

ANSWER

Let the charge is placed at distance x from 2μc

Now the net force acting on charge q should be zero.

⇒

(10−x)

2

k(1×10-6)q =2(2*10-6)q

⇒

10−x1=2

⇒x=10

2-x2

⇒x(1+2)=10

2⇒x= 1+ 210

2⇒x=

2.414

14.14

⇒x=5.857cm

Explanation:

Hope it helps you!