Finding Charge & Tension of a Suspended Cork Ball

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SUMMARY

The discussion focuses on calculating the charge and tension of a suspended cork ball in a uniform electric field defined by E=(3×10^5 N/C)i+(5×10^5 N/C)j. The equilibrium conditions lead to the equations Tsin(37)=(3*10^5)q and Tcos(37)+(5*10^5)q=mg, where the mass of the ball is 0.7g and gravity is 9.8 m/s². The final calculated charge on the ball is 7.6382 nC, which should be rounded to 8 nC due to significant figures, and the tension in the string is determined to be 0.003808 N.

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Homework Statement


A charged cork ball is suspended on a light string in the presence of a uniform electric field.
E=(3×10^5 N/C)i+(5×10^5 N/C)j. The ball is in equilibrium in the field.
Find the charge on the ball. The acceleration due to gravity is 9.8 m/s2 . The mass of the ball is 0.7g. The length of the string is 1.4m and at an angle of 37 degrees from the center.
Answer in units of nC.

Also find the tension of the string.

Homework Equations


Tsin(theta)=FEx
Tcos(theta)+FEy=mg
F=Eq

The Attempt at a Solution


Will someone please let me know if my solution and answer is correct?

Tsin(37)=(3*10^5)q
Tcos(37)+(5*10^5)q=(.0007)(9.8)
T=(3*10^5)q/sin(37)
(3*10^5)q*cot(37) + (5*10^5)q = .00686
Solve for q and q=7.6382nC

T=(3*10^5)q/sin(37) plug in q and T=.003808N
 
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It looks good. You might have put in some comments regarding what you were accomplishing or doing at each step in order to make your thought process clear. Otherwise, very well done!
 
@gneill Thanks! I will definitely do so next time :)
 
amyc said:
q=7.6382nC
Several of the given numbers are only to one significant figure, so technically you should give the answer as 8nC.
 

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