(a) What is the magnitude of the net electric field where the man is standing? N

Question

(a) What is the magnitude of the net electric field where the man is standing?
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.)
? counterclockwise from the +x-axis

Please help thank you A man is standing beneath two rain clouds, equidistant from each cloud, as shown in the figure. Cloud A has an excess of 4.0 (a) What is the magnitude of the net electric field where the man is standing? 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.) ° counterclockwise from the +x-axis 10^4 positive charges (where a charge has magnitude e). 10^4 negative charges while cloud B has 4.0

Explanation / Answer

1) the negative charges and positive charges are equal then q=ne

q=104*4*1.6*10^-19C

=665*10^-19 C

The electric field of the man standing is

E=(1/4*pi*epsilon not)(q1*q2/r^2)(sin(theta) in direction of x axis

But the vertical component can be canceled

changing the expression

E=(1/4*pi*epsilon not)(q*d/((z^2+(0.25)^2)^(3/2))

then substitute the values in above expression,

=(1/4*3.14*8.85*10^-12)((665*10^-19)^2*(0.5))/((0.25)^2+(2)^2)^(3/2))

=352.39*10-13N/C

2) The net electric field acting on the direction of z axis.

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