Calculating Potential Difference Between A & B

In summary, the conversation is about finding the potential difference between two points A and B in space using the definition of potential difference, E = -\nabla\phi. The attempt at a solution involves using the del operator and the formula V = \int{^A_B}{E\cdot{dr}} to find V, but the proposed solution does not make sense. The expert encourages the person to reconsider their approach and suggests calculating \vec{E}\centerdot d\vec{r} in terms of \phi.
  • #1
girlinphysics
25
0

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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  • #2
You do know that V and φ are the same parameter, correct?

Chet
 
  • #3
Yes I do know that.
 
  • #4
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?
 
  • #5
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
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential between two points in a circuit. It is measured in volts (V) and is a measure of the electrical energy needed to move a unit of charge from one point to another.

2. How do you calculate potential difference?

Potential difference can be calculated by dividing the work done by the charge moved. It can also be calculated by multiplying the current (I) in a circuit by the resistance (R) in ohms (Ω), using the formula V=IR.

3. What factors affect potential difference?

The main factors that affect potential difference are the strength of the electric field, the amount of charge being moved, and the distance between the two points. Temperature and the material of the conductor can also have an impact on potential difference.

4. How does potential difference affect current?

Potential difference is directly related to current, as stated in Ohm's Law (V=IR). This means that as the potential difference increases, the current also increases, and vice versa. However, the resistance in the circuit can also affect the current.

5. What are some real-life applications of potential difference?

Potential difference is an important concept in many electrical devices and systems, such as batteries, power lines, and electronic circuits. It also plays a role in many everyday activities, such as charging a phone or using household appliances. Understanding potential difference is crucial in designing and maintaining safe and efficient electrical systems.

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