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 4.5 x 104 negative charges while cloud B has 4.5 x 104 positive charges (where a charge has magnitude e) 0.5 km 2 km (a) What is the magnitude of the net electric field where the man is standing? 7.4E-12 What is the equation for the electric field from a point charge? How do you determine the net electric field from multiple charges? N/C (b) What is the direction of the net electric field where the man is standing? (Assume the +x-axis is to the right.) 180 °Counterclockwise from the +x-axis

Explanation / Answer

Magnitude of Ef due to A and B = kq/r^2 = 9*10^9*4.5*10^4 / (2^2 + 0.25^2)*1000

= 10^10 N

for cloud A , direction will be towards A and for cloud B direction will be be away from B

Net EF in Vertical direction will be same and cancel out

Net EF in x-direction = 2*10^10*0.25 / sqrt(0.25^2 + 2^2)

= 2*10^9 N /C

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