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Electric Fields and Potential Difference

  1. May 9, 2009 #1


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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?

  2. jcsd
  3. May 9, 2009 #2


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    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]?

    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.

    After the first question, you should be in a position to tell me :wink:
  4. May 9, 2009 #3


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    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.

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