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Potentials and voltages

1. Homework Statement

Point X is 0.25m away from a point charge of +4.7 E -8 C, point Y is 0.65m away. What is the potential of point X with respect to point Y?

2. Homework Equations
delta Ep=delta Ek
V=delta Ep/Q
delta Ep=QV
Ep=kQQ/d

3. The Attempt at a Solution

delta Ep = -Ep1
W=QV
W=kQQ(1/d-1/d)
QV=kQQ(1/.25-1/.65)
V=(9E9)(4.7E-8 C)(2.4615)
V=1041 V

I don't get one thing when it says with respect to Y, you put the distance of Y as d2 right? But then wouldn't you get a negative answer when it should be positive?
 

Hootenanny

Staff Emeritus
Science Advisor
Gold Member
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delta Ep = -Ep1
W=QV
W=kQQ(1/d-1/d)
QV=kQQ(1/.25-1/.65)
V=(9E9)(4.7E-8 C)(2.4615)
V=1041 V
What you have done here is all correct, except the answer. You've just punched the numbers into your calculator wrong.
I don't get one thing when it says with respect to Y, you put the distance of Y as d2 right? But then wouldn't you get a negative answer when it should be positive?
The potential of x wrt to y means exactly what you have done there. In other words calculate the potential difference between points x and y like this; [itex]V_{xy} = V_{y}-V_{x}[/itex], which would lead to a positive answer.
 

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