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

Question

Quizplus · Accepted Answer

The electric field due to a point charge at a distance r from the charge is given by kQ/r^2 where k is Coulomb's constant. 
At point P, the distance from the charge Q is d = √(a^2 + b^2) = √(3^2 + 4^2) = 5 m.
Therefore, the electric field at point P is E = kQ/d^2 = (9x10^9 Nm^2/C^2)(16x10^-9 C)/(5m)^2 = 23.04 N/C, which is closest to option C (24 N/C).

If Q = 16 NC,a = 3