Calculating Potential Difference Between A & B

girlinphysics
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Homework Statement


Two points A and B are separated in space by distance dr.
A has coordinates (x,y,z) and B has coordinates (x+dx, y+dy, z+dz).
Using the definition of potential difference, show [itex]E = -\nabla\phi[/itex]

Homework Equations


[itex]E = -\nabla\phi[/itex]
[itex]V = \int{^A_B}{E\cdot{dr}}[/itex]

The Attempt at a Solution


With the two formulas listed above, I think I can find V, taking [itex]E = (\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex] from the question which means simply [itex]V = (x,y,z)[/itex]. Since the del operator is [itex](\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex], applying that to V you get [itex]E = (\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex].

I know its not this simple because it seems like my maths has gone around in circles. Any help please?
 
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You do know that V and φ are the same parameter, correct?

Chet
 
Yes I do know that.
 
girlinphysics said:

Homework Statement


Two points A and B are separated in space by distance dr.
A has coordinates (x,y,z) and B has coordinates (x+dx, y+dy, z+dz).
Using the definition of potential difference, show [itex]E = -\nabla\phi[/itex]

Homework Equations


[itex]E = -\nabla\phi[/itex]
[itex]V = \int{^A_B}{E\cdot{dr}}[/itex]

The Attempt at a Solution


With the two formulas listed above, I think I can find V, taking [itex]E = (\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex] from the question which means simply [itex]V = (x,y,z)[/itex]. Since the del operator is [itex](\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex], applying that to V you get [itex]E = (\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})[/itex].

I know it's not this simple because it seems like my maths has gone around in circles. Any help please?
What you wrote doesn't really make sense. For one thing, potential is a scalar, so claiming V=(x,y,z) is nonsense. And what is ##E = (\frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz})## supposed to mean?
 
girlinphysics said:
Yes I do know that.
So, when you calculate ##\vec{E}\centerdot d\vec{r}## in terms of ##\phi##, what do you get?

Chet
 

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