Potential Difference between 2 points

AI Thread Summary
The discussion focuses on calculating the potential difference between two points, A and B, influenced by a +4.0 μC and a -4.0 μC point charge. Initially, the potential at each point was incorrectly calculated by only considering the contribution from one charge. The correct approach involves summing the potentials from both charges at each point, leading to a potential of 48,000 V at point A and -48,000 V at point B. This results in a potential difference of 96,000 V between points A and B. The final conclusion confirms the accurate calculation of the potential difference.
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Homework Statement


A +4.0 μC-point charge and a -4.0 μC point charge are placed as shown in the figure. What is the potential difference, V_A - V_B, between points A and B? (k = 1/4πε_0 = 8.99 × 10^9 N · m^2/C^2)

https://www.physicsforums.com/attac...2/?temp_hash=fc256c7e6e7317423fc360366b1c7272

Homework Equations


V = (k*q)/r

The Attempt at a Solution


Point B is 0.30 m away from the -4.0 μC point charge. The potential there is (9*10^9*-4*10^(-6))/(.3) = -120,000 V.
Point A is 0.30 m away from the 4.0 μC point charge. The potential there is (9*10^9*4*10^(-6))/(.3) = 120,000 V.
So I thought I could do V_A - V_B = 120,000V - (-120,000V) = 240,000V but that doesn't work.
Help please? Thanks.
 

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The point here is to create the algebraic sum of potential differences between points A and B. So, what about between the two charges?
 
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You need to account for the potentials due to both charges at both locations. So for example the potential at point A is due to the effects of both the 4 μC charge and the -4 μC charge combined.
 
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So the potential at A is:
(9*10^9*4*10^(-6))/(.3) + (9*10^9*-4*10^(-6))/(.5) = 48,000 V
and the potential at B is:
(9*10^9*-4*10^(-6))/(.3) + (9*10^9*4*10^(-6))/(.5) = -48,000 V
V_a - V_b = 96,000 V?
 
That looks better.
 
Thanks! Got it.
 
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