How Do You Calculate the Electric Potential Difference Between Two Points?

[Fusilli_Jerry89]

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?

Homework Equations

delta Ep=delta Ek
V=delta Ep/Q
delta Ep=QV
Ep=kQQ/d

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]

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; $V_{xy} = V_{y}-V_{x}$, which would lead to a positive answer.

 

[m2003]

I appreciate your attempt at solving this problem using the given equations. However, there are a few things that could be clarified and improved upon in your solution.

Firstly, when calculating the potential at point X with respect to point Y, you would use the distance between the two points (0.25m) as the value for d in the equation Ep=kQQ/d. This is because the potential at a point is only dependent on the distance from the point charge, not on the distance from another point.

Secondly, in your calculation for V, you have used the equation W=kQQ(1/d1-1/d2), which is the formula for the work done by an electric field. This is not the same as the potential difference (V) between two points, which is given by the equation V=kQ/d. Additionally, you have used the distance between the two points (0.25m) as d1, which would give a negative value for V as the distance between two points is always positive.

To calculate the potential at point X with respect to point Y, you would use the equation V=kQ/d, where d is the distance between the two points (0.25m). This would give a potential of approximately 1.04 x 10^3 V, or 1040 V. This value is positive, as expected.

In conclusion, the potential at point X with respect to point Y is approximately 1040 V. It is important to use the correct equations and values when solving physics problems to ensure accurate results.