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# Need help with physics! A man is standing beneath two rain clouds, equidistant from each cloud...?

A man is standing beneath two rain clouds, equidistant from each cloud, as shown in the figure. Cloud A has an excess of 6.6*10^4 negative charges while cloud B has 6.6* 10^4 positive charges (where a charge has magnitude e).

Cloud A and B are 0.5 km apart and 2 km above the man (I don’t know how to insert the diagram, sorry!)

WHAT IS THE MAGNITUDE OF THE NET ELECTRIC FIELD WHERE THE MAN IS STANDING?

### 1 Answer

- RealProLv 71 year agoFavorite Answer
Have you tried using the formulas?

When a point charge Q is at a distance x horizontally and y vertically from a point, then r^2 = x^2 + y^2

so the field has magnitude E = kQ / r^2

But to get the horizontal and vertical components of field, we have to use similar triangles

In horizontal direction, E_x = E * x/r

= k*Q*x / r^3

Because of symmetry, vertical components of field will cancel out and the horizontal ones will add up.

Total E = 2E_x

= 2k*Q*x / r^3

= 2k*(6.6×10^4 e) * (250 m) / (250^2 + 2000^2 m^2)^(3/2)

= 5.8×10^-12 N/C

Source(s): Clouds that close are hardly points- Login to reply the answers