What are the strength and direction of the electric field at the position indica

Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure where q = 1.9 nC? Specify the direction as an angle above (positive) or below (negative) horizontal (the positive x-axis).

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What are the strength and direction of the electric field at the position indicated by the dot in the figure below, in which d = 4 cm and q = 2 nC? Specify the direction as an angle measured counterclockwise from the positive x-axis, which points to the right.

________N/C
_________ °

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q = ________________ nC

Explanation / Answer

For +1.9 nC charges, kQ = 8.99*10^9 * 1.9*10^-9 = 17.08

Electric field strength E = kQ / r^2
Electric field direction is radially away from + charges

At the dot:
r1 = 5cm = 0.05m
so E1 = 17.08 / (0.05)^2 N/C = 6832.4 N/C @ direction angle 0.
E1 = 6832.4i + 0j

r2 = sqrt(5^2 + 10^2) cm = 11.18 cm = 0.112 m)
so E2 = 17.08 / (0.112)^2 = 1361.6 N/C
Angle is 90 - arctan(5/10) = 63.43 deg
E2 = 609.03i + 1217.8j

So the net field strength at the dot is
E = E1 + E2 = 7441.4i + 1217.8j = 7540.3 N/C @ 9.29 deg above horizontal.

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