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3 years ago

If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?

1 answer:

8 0

In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C

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