Electric Potential Energy of a Charge Distribution

AI Thread Summary
The discussion focuses on calculating the total electric potential energy and potential at a specific point for a system of three charged spheres. The total electric potential energy (EE) was found to be -405 J using the formula EE = k(q1q2/r + q2q3/r + q1q3/r). The electric potential (V) at point Z was calculated to be approximately 3.7 x 10^5 V. Participants noted the importance of careful calculation and the use of brackets in equations to avoid errors. Overall, the calculations were confirmed by multiple contributors, indicating consistency in the results.
HarleyM
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Homework Statement


Examine the diagram provided,
Sphere 1: q=3.0x10-4C
Sphere 2: q=-3.0x10-4C
Sphere 3: q= 3x10-4C
digram.jpg

a) Find the total electric potential energy of the charge distribution
b) Find the total electric potential at point Z

Homework Equations


EE=kq1q2/r +kq2q3/r+kq1+q3/r... etc

V=kq/r

The Attempt at a Solution



a) EE= kq1q2/r +kq2q3/r+kq1+q3/r
EE=(9x109)[(3.0x10-4*(-3.0x10-4)/3) + ((-3.0x10-4*(3.0x10-4)/3) + ((3.0x10-4*3.0x10-4)/6)]
EE= -405 J (Does anyone else get the same answer? I tried to do it continuously in my calculator so I don't lose digits)

b) V= K(q1/r+q2/r+q3/r)
V=(9x109)[(3.0x10-4/4.24)+(-3.0x10-4/3)+(3.0x10-4/4.24)]

V=3.7x105 V or J/C
 
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Those answers look good to me.
 
Ok thanks!
 
Those are tough to calculate all at once. It took me a few tries but I did end up with the same answers.
For anyone else who likes to take things one step at a time, here you go:

a) E = k[(-3x10^-8)+(1.5x10^-8)+(-3x10^-8)]

b) V = k(4.15094x10^-5)

(and at the end, when your multiplying the final sum by Coulomb's Constant (k=9x10^9), make sure to put brackets around that too. Seems to make a difference.)
 
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