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

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SUMMARY

The discussion centers on calculating the electric potential at the origin due to four point charges located at specified coordinates. The charges are 12μC at (-4,4) and (4,4), -6μC at (-4,-4), and -3μC at (4,-4). The correct formula for the distance 'r' was identified as 4√2, leading to the correct potential calculation of 47676.7 V after adjusting the divisor. The user resolved their error by recognizing the incorrect division by 2, confirming the importance of accurate distance measurement in potential calculations.

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Homework Statement



4 charges, distributed as follows

12*10[itex]^{-6}[/itex] C @ (-4,4)
12*10[itex]^{-6}[/itex] C @ (4,4)
-6*10[itex]^{-6}[/itex] C @ (-4,-4)
-3*10[itex]^{-6}[/itex] 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



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


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


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
 
Last edited:

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