Homework Help: Calculate the charge on the ball

waywardtigerlily · Jan 25, 2005

I have been working on this problem all night. I am not looking for a numberical answer, just help with the equations. A charged cork ball of mass 1.15 g is suspended on a light string in the presence of a uniform electric field.

When the electric field has an x component of 3.05E+5 N/C, and a y component of 4.81E+5 N/C, the ball is in equilibrium at q = 35.7o. Calculate the charge on the ball.

I can not figure out how to find the electric field using the x and y components. One I figure that out I can get the problem.

Any help would be appreciated.
Thanks!

vincentchan · Jan 25, 2005

there are 2 forces acting on the ball
1. electrostatic force
2. gravity
draw the free body diagram, and realize Fx/Fy=tan35.7

waywardtigerlily · Jan 25, 2005

I still can't figure it out..I have q=(mgtan)/ E What I can't figure out is E. I know how to work the problem in reverse, when Given E and having to find Fx and Fy. Then finding the Magnitude (E= sqrt(fx^2+Fy^2)).

apchemstudent · Jan 25, 2005

waywardtigerlily said:

I still can't figure it out..I have q=(mgtan)/ E What I can't figure out is E. I know how to work the problem in reverse, when Given E and having to find Fx and Fy. Then finding the Magnitude (E= sqrt(fx^2+Fy^2)).

I don't know why you're trying to solve E as sqrt(fx^2+Fy^2)... It's actually very useless to solve the problem that way.

Why don't you just solve the components of the forces seperately since you are given E for each? Keep in mind that you will just have to keep the charge as a variable...

waywardtigerlily · Jan 25, 2005

do you mean: Ex=(kq/r^2)cos and Ey=-(kq/r^2)sin...if so I don't know how to get the radius without the length of the string.

apchemstudent · Jan 25, 2005

waywardtigerlily said:

do you mean: Ex=(kq/r^2)cos and Ey=-(kq/r^2)sin...if so I don't know how to get the radius without the length of the string.

Ex = Fx/q where q is the charge

Fx = Ex*q <----- you can use this as the substitute for the Force in x component

I think you can take it from here...