Question

Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle, as the drawing shows. The electric field at the midpoint M between the charges has a magnitude E_M. The field directly above the midpoint at point P has a magnitude E_P. The ratio of these two field magnitudes is E_M/E_P = 6.29. Find the angle ? in the drawing.

Answer 1

The first thing I would do is find the electric field at both the points M and P. I'm going to call the distant from a charge to the midpoint d and the distance between a charge and the point p, r. Since both of the point charges are of opposite charge, their magnitudes should add.

So, at point M, the field should be:
Em = (kq)/d^2 + (kq)/d^2 = (2kq)/d^2

At point P, the field should be:
Ep = (kq)/r^2 + (kq)/r^2 = (2kq)/r^2

Since they give you the ratio and now you have the expressions for the fields, you now have an equation:
Em/Ep = 6.29 = (2kq)/d^2 * r^2/(2kq) = r^2/d^2

Now, looking at the triangle, you can say that the angle between the base and the point P can be cos(alpha) = d/r

From the previous equation, we can find that r/d = 2.51. So, d/r = 1/2.51.

This means that the angle should be arccos(1/2.51) which is about 66.521 degrees.