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
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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
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