4 charges, find potential


by jehan60188
Tags: charges, potential
jehan60188
jehan60188 is offline
#1
Feb1-12, 03:33 PM
P: 204
1. The problem statement, all variables and given/known data

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.


2. Relevant equations

V= U/q = -W/q = ∫E*dl = k*q/r

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


3. 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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tiny-tim
tiny-tim is offline
#2
Feb1-12, 03:47 PM
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hi jehan60188!
shouldn't r be 4√2 ?
jehan60188
jehan60188 is offline
#3
Feb1-12, 05:46 PM
P: 204
Quote Quote by tiny-tim View Post
hi jehan60188!
shouldn't r be 4√2 ?

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!

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