Question

two point charges Q1 equals to 400 microcoulomb and Q2 = 100 microcoulomb are kept 60 CM apart in vacuum find intensity of the electric field at midpoint of the line joining Q1 and Q2​

Answer 1

this is your answer!!!!!

Answer 2

Explanation:

Let E1 be the electric field due to charge Q1 and E2 be the electric field due to charge Q2.

Then the electric fields are given as :

E1 = 9 × 10^9 × C/(d/2)²

Where C = 400mC/1000 = 0.4 C

d/2 = 60/2 = 30cm = 30/100 = 0.3 m

E1 = 9 × 10^9 × 0.4/0.3² = 4 × 10^10 N/C

E2 = 9 × 10^9 × 0.1/0.3² = 1 × 10^10 N/C

The effective electric field acting at the midpoint (Em) is given by :

Em = E1 + E2

Em = 1 × 10^10.+ 4 × 10^10 = 5 × 10^10 N/C