Question

Two charges of 2 into 10 raise to minus 8 coulomb and minus 10 raise to minus 8 coulomb are placed at point A B respectively find the resultant force on a charge of 1 microcoulomb held at a point such that ap is equal to 10 BP is equal to 5 and angle APB is equal to 90 degree

Answer 1

Thus the values of electric field due to a ad B are given as below.

Explanation:

• Electric field due to A.

EA = kq /10^2 × 10^−4 = 9× 10^9 × 2× 10^−8 / 10^2 × 10^−4 = 18000 v/m

• Electric field due to B.

EB  = kq 10^2 × 10^−4 = 9 × 10^9 × 10^−8 / 25 × 10^−4 = 36000 v/m

Resultant electric field:

E = √ E^2A + E^2B

E = √(18000)^2 + (36000)^2 = 4024.9 v/m

F = q.E = 1 × 10^−6 × 4024.9 = 4.024 × 10^−3 N

Direction would be tanθ = Eb / EA = 36000 / 18000 = 2