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?
- 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/CSource(s): Clouds that close are hardly points