What Is the Electric Potential at the Origin Given Four Point Charges?

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The discussion focuses on calculating the electric potential at the origin due to four point charges with specified magnitudes and positions. The initial attempt at the solution involved summing the potentials from each charge but yielded an incorrect result. A key correction was identified regarding the distance used in the calculations, specifically that the correct distance should be 4√2 instead of 2. After adjusting this error, the user successfully arrived at the correct answer. The thread concludes with a request for moderators to close the topic.
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Homework Statement



4 charges, distributed as follows

12*10^{-6} C @ (-4,4)
12*10^{-6} C @ (4,4)
-6*10^{-6} C @ (-4,-4)
-3*10^{-6} C @ (4,-4)

Calculate the potential at the origin if the potential at infinity is zero.


Homework Equations



V= U/q = -W/q = ∫E*dl = k*q/r

for multiple point charges, find V for each one, and sum them up


The Attempt at a Solution



V1 = V2 = \frac{k*12*10^{-6}}{(4*\sqrt{2}/2)}
V3 = \frac{k*-6*10^{-6}}{(4*\sqrt{2}/2)}
V4 = \frac{k*-3*10^{-6}}{(4*\sqrt{2}/2)}


V1+V2+V3+V4 = \frac{k*3*10^{-6}}{4*\frac{\sqrt{2}}{2}}*(4+4-2-1) = 47676.7<br />


So, I factor out the k, the 3*10e-6, and the 1/(r) from each Vi, then multiply it by 4+4-2-1 =5

this answer is not right. Any pointers?

thanks!]
 
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hi jehan60188! :smile:

shouldn't r be 4√2 ? :redface:
 
tiny-tim said:
hi jehan60188! :smile:

shouldn't r be 4√2 ? :redface:
4^2 + 4^2 = 32 = 4 Sqrt(2)
why, WHY did I divide by 2?!

divided my solution by 2 (since it was sqrt(2)/2 in the divisor), and got the right answer! thanks!

mods, please feel free to close topic
 
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