Calculate the charge on the ball

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To calculate the charge on a cork ball suspended in an electric field, the discussion emphasizes the need to analyze the forces acting on the ball, specifically the electrostatic force and gravity. The equilibrium angle of 35.7 degrees allows for the use of trigonometric relationships to relate the forces. Participants suggest focusing on the components of the electric field rather than calculating its magnitude using the Pythagorean theorem. The equations Ex = Fx/q and Fx = Ex*q are highlighted as useful for substituting variables to solve for charge. Understanding these relationships is crucial for progressing in the problem.
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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!
 
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there are 2 forces acting on the ball
1. electrostatic force
2. gravity
draw the free body diagram, and realize Fx/Fy=tan35.7
 
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)).
 
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...
 
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.
 
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...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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