Please show how you got answers Calculate the net electric field (magnitude and

Question

Please show how you got answers

Calculate the net electric field (magnitude and direction) at point A in the diagram of charges shown on the right. Each charge has a magnitude of 1 nC and each small division is 0.2m 1. The electric field due to a point charge q1 is E=ref where Nm c2 distance between the charge and point A, and f is a unit vector directed either away or towards q1. The net electric field is the vector sum of the electric fields of each charge (account for direction) Figure for questions #1 & 2

Explanation / Answer

1)The field along x will be due tothe two negative charges.

Since the magnitudes of the charges are same, so is the sign and distances from point A. The field in x direction will get cancelled because of the equal magnitude but apposite direction.

Ex = 0

The field in y direction will be:

Ey = k q/ry^2 = 8.99 x 10^9 x 1 x 10^-9/2^2 = 2.25 N/C

total electric field at A will be:

Ea = sqrt(Ex^2 + Ey^2) = 2.25 N/C

Hence, Ea = 2.25 N/C ; Direction will be downwards.

2)The electric potential at A will be the sum of individual potentials due to the charges

V = V1 + V2 + V3

V = k q (1/r1 + 1/r2 + 1/r3)

V = 8.99 x 10^9 x 1 x 10^-9 (-1/1 -1/1 + 1/2) = -13.5 V

Hence, V = -13.4 V

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