ELECTRICAL FIELD, Please help!?

Question

Three negative charges (all similar) of q are 5microcoulombs. These lie on a circle with a radius of 2.m at angles 30, 150 and 270. What is the resultant electrical field in the center of this circle?

Answer 1

I just did the math, zero is indeed correct.

If you ever do problems with vectors all you need is moxy.

M for magnitude O for degrees X for m*cos(degrees) and Y for M*sin(degrees).

In this case

M O X Y
5 150 5*cos(150) 5*sin(150)
5 30 5*cos(30) 5*sin(30)
5 270 5*cos(270) 5*sin(270)
_____________________________
0 0 the sum

if the sum is non zero for either X or Y, draw it out on a circle to get the correct value for degrees.

So here is an example
x = .5 and y = -.5

If you draw out a circle and draw the line that corresponds to these data, you find that the angle is in the second quadrant.

Make both the X and Y terms positive, (take their absolute value) and plug them in thusly:

arctan(y/x) = whatever degree value (which is the angle that the line subtends to the nearest axis).

Answer 2

The electric field from each charge is E = q/4*π*e0*r^2. By symmetry, the resultant field will be along the y-axis. The result, from geometry

E - 2*E*sin(30º) = E- 2*E*.5 = E-E = 0

The first E is upward, from the charge at 270º, the other two downward from the y-component of field from the other two charges, both equal to E*sin(30º).

The horizontal field components of the charges at 30º and 150º are equal and opposite so they also cancel. The charge at 270º produces no horizontal component at the origin.

Answer 3

Um... I think it is zero. I may be making a mistake but I believe they all cancel out since they are evenly spaced on a circle.