A man is standing beneath two rain clouds, equidistant from each cloud, as shown

Question

A man is standing beneath two rain clouds, equidistant from each cloud, as shown in the figure. Cloud A has an excess of 4.4 X 10^4 negative charges while cloud B has 4.4 X 10^4 positive charges (where a charge has magnitude e). (a) What is the magnitude of the net electric field where the man is standing? What is the equation for the electric field from a point charge? How do you determine the net electric field from multiple charges? (b) What is the direction of the net electric field where the man is standing? (Assume the +x - axis is to the right.)

Explanation / Answer

The field from cloud A points toward A, while the cloud from cloud B points away from B.
distance d = (0.25² + 2²) km = 2016 m
E = kQ / d², so
Ea = (8.99x10^9 N·m²/C² )x (4.4 x10^4) x(1.6x10^-19C) / (2016m)² = 1.55x10^-11 N/C
and angle a = arctan(2/-0.25) = -83º + 180º = 97º from +x axis
Eb = 2.73e-11 N/C
at b = 83º + 180º = 263º from +x axis

Due to symmetry, the vertical components cancel and the net field is to the left with magnitude
E = 2 * 1.55x10^-11N/C * cos83º =3.77x10^-12 N/C to the left
(180º from +x axis)

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