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Potential Difference between two positive point charges 
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#1
Apr1813, 08:30 PM

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1. The problem statement, all variables and given/known data
Two equal positive charges Q are fixed on the xaxis, one at +a and the other at a. (a) The electric field E at the Origin O (b) The electric potential V at the origin O 2. Relevant equations E=dV/dr > V=kQ/r 3. The attempt at a solution V_{Net} = V_{0} + V_{1} I got V_{0} = kQ/r (because the vector r is negative) ..and V_{1} = kQ/r V = 0 Is this correct? Thank you for your help! 


#2
Apr1813, 09:11 PM

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#3
Apr1813, 09:14 PM

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Hi Flannabhra!
Let's answer a few questions: a.) What kind of quantity is the electric potential, V? b.) Describe to me the significance of 'r'. 


#4
Apr1813, 09:16 PM

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Potential Difference between two positive point charges
First of all, what was your answer for part (a) and how did you get it. For you attempt at part (b): What does the variable, r, represent? 


#5
Apr1813, 09:24 PM

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#6
Apr1813, 09:52 PM

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#7
Apr1813, 09:58 PM

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On the matter of r, I was under the impression that you could take the direction that r is pointing from the reference point and use that as the direction for r. I.e. the equation for E would be: E = kQ/r^{3} * r  in which r is the vector r and the r in the denominator is raised to the third power so as to keep the equation true. 


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Apr1813, 10:33 PM

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#9
Apr1813, 10:42 PM

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#10
Apr1813, 10:44 PM

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a) I believe that in this situation, V is scalar. b) 'r' is the distance from the point Q to the origin 


#11
Apr1813, 11:03 PM

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For the charge at x = a, r = ai . In each case the distance is a, and the vector, r, points away from the charge. 


#12
Apr1813, 11:08 PM

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Thank you so much for your help SammyS!!! 


#13
Apr1813, 11:57 PM

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