Electric Fields and Potential Difference

In summary: After the first question, you should be in a position to tell me :wink:The potential difference at any point inside of a charged conducting sphere is the same as the potential difference at any point inside of a charged conducting cube.
  • #1

jgens

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Just a few brief questions that my teacher did not answer during the last class:

1) Why is the potential difference at any point inside a charged conducting sphere the same?

2) If a charge were placed outside of a charged say, conducting cube, would the electic field drop off proportionally with r2?

3) Would the potential difference at any point inside of a charged concucting cube be the same as with a sphere?

Thanks!
 
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  • #2
jgens said:
Just a few brief questions that my teacher did not answer during the last class:

1) Why is the potential difference at any point inside a charged conducting sphere the same?

You seem to be confusing "potential difference" with "electrostatic potential" (unless it was just a typo): Electrostatic potential [itex]V(\vec{r})[/itex] has a defined value at any given point, while "potential difference" refers to the difference in potential between two different points (i.e. [itex]V(\vec{r}_2)-V(\vec{r}_1)[/itex] ) and so saying "the potential difference at any point..." makes no sense.

As for why the electrostatic potential has the same value at any point inside a conductor (charged or otherwise, it doesn't matter), go back to the definition of electrostatic potential:

[tex]V(\vec{r})=-\int_{\mathcal{O}}^{\vec{r}} \vec{E}(\vec{r})\cdot d\vec{l}[/tex] Where [itex]\mathcal{O}[/itex] represent the chosen reference point (usually infinity)

Or, better yet, look at the potential difference between any two points inside the conductor:

[tex]V\equiv V(\vec{r}_2)-V(\vec{r}_1)=-\int_{\vec{r}_1}^{\vec{r}_2} \vec{E}(\vec{r})\cdot d\vec{l}[/tex]

You should be aware that this integral is path independent, and hence you are allowed to evaluate it using any path that connects the two points [itex]\vec{r}_1[/itex] and [itex]\vec{r}_2[/itex]...What happens if you chose a path that is entirely inside the conductor? Specifically, what is the value of [itex]\vec{E}(\vec{r})[/itex] inside the conductor, and what does that make [itex]V[/itex]?

2) If a charge were placed outside of a charged say, conducting cube, would the electic field drop off proportionally with r2?

Far away from the conductor, yes. However, the point charge would induce a non-zero charge density on the conductor (if the point charge is positive, positive charges in the cube will be repelled away from the point charge and negative charges will be attracted closer to it). This induced charge density will alter the electric field close to the conductor so that it will not, in general, fall off like 1/r^2. Far away from the conductor, the field will look approximately like that of just a point charge and will fall off like 1/r^2.


3) Would the potential difference at any point inside of a charged concucting cube be the same as with a sphere?

Thanks!

After the first question, you should be in a position to tell me :wink:
 
  • #3
Thanks for catching my terminology error gabbagabbahey. Unfortunately, I'm currently enrolled in an algebra-based physics course (only physics course my high school offers) so I'm not familiar with that definition of electrostatic potential. Also, while I am comfortable with calculus, I would like to try and understand these problems using the concepts developed in my class (I need to elaborate more, I know). However, thank you for your well thought-out response.

Suppose I'll try again:

1) I thought that the electrostatic potential inside any given point of a conductor would be zero because, in my physics class, we defined electrostatic potential as electric potential energy per unit charge (V = E*h). Since the electric field inside the conductor is zero, I thought that V = 0; however, my physics teacher said that the electrostatic potential equals some non-zero value.

2) I understand your explanation very well, thank you!

3) I'm assuming if I get the first I'll be able to get this one.

Thanks!
 

What is an electric field?

An electric field is a region around a charged particle or object where other charged particles will experience a force. This force is either attractive or repulsive depending on the charge of the particle.

How is an electric field created?

An electric field is created by a charged particle or object. When a charged particle is placed in a region, it will create an electric field around it. The strength of the electric field depends on the magnitude of the charge and the distance from the particle.

What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric field. It is measured in volts and is the driving force for the flow of electric current.

How is potential difference related to electric field?

The potential difference between two points in an electric field is directly proportional to the strength of the electric field and the distance between the two points. This relationship is described by the equation V = Ed, where V is potential difference, E is electric field strength, and d is distance.

What is the unit of measurement for electric field?

The unit of measurement for electric field is newtons per coulomb (N/C). This unit represents the force per unit charge that a charged particle would experience in an electric field.

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