Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The question asked me to find the potential at a distance $r$ from the center of a charged sphere, where $r>r_0$ of the sphere.

Actually, the question is answered, but what is confusing me is that the author took 2 points out of the sphere ($a$ & $b$) and calculated the potential difference between them, and getting the result that $$V= \frac{Q}{4\pi \epsilon_0 r} $$ at any point outside the sphere?

Why did he take these 2 points ?

share|improve this question
    
When asking questions on physics.stackexchange, you can use symbolic math. For example, your last equation can be written \$V=Q/(4 \pi \epsilon_0 r)\$, which will produce the output $V=Q/(4 \pi \epsilon_0 r)$. –  KDN Mar 2 '13 at 16:25
    
$V=Q/(4 \pi \epsilon_0 r)$ yeh it worked I don't know about Abdulrahman Hessen but it solve my problem –  Akash Mar 2 '13 at 17:29
add comment

2 Answers

up vote 1 down vote accepted

Potentials are curious in that there isn't exactly one way to define them. Because a potential $\phi$ is related to a field $\mathbf{E}$ by \begin{equation} \mathbf{E}=\nabla \phi \end{equation} the class of potentials that satisfy the field equation is infinite. If a potential $\phi$ satisfies the field equation, then so will any potential $\phi'$, where \begin{equation} \phi'=\phi+\Lambda \end{equation} where $\Lambda$ is any function that satisfies $\nabla \Lambda=0$. The simplest example is for $\Lambda$ to be some constant value, but in general, an even larger class of functions works.

By taking a difference of potentials, there is no ambiguity. If the potential at point $a$ and the potential at point $b$ are defined in terms of $\phi$, their potential difference is \begin{equation} \Delta \phi = \phi(b)-\phi(a). \end{equation} Doing this same exercise in terms of $\phi'$ gives \begin{equation} \Delta \phi' = \phi'(b)-\phi'(a)=\left(\phi(b)+\Lambda \right)-\left(\phi(a)+\Lambda \right) = \phi(b)-\phi(a)=\Delta \phi. \end{equation} So while potentials are potentially ambiguous, differences in potentials are not. By convention, physicists frequently assign a potential of "$0$" to a point at infinity. This fixes the value of $\Lambda$ and removes the ambiguity in the definition of potential. This is only convention, however, and there are certain problems that this approach cannot resolve. By defining the potential in terms of differences, your author avoided the issue entirely.

share|improve this answer
    
Thank you vey much ^^ –  Abdulrahman Hessen Mar 2 '13 at 18:47
1  
@Rafael: nicely edited. I wish there were a way to upvote an edit. –  KDN Mar 3 '13 at 0:31
add comment

Well usually one computes the Electric field using the Gauss Theorem such that

$$\oint_{S} \vec{E}\cdot \vec{n}dS=\frac{q_I}{\epsilon_0},$$

where $q_I$ is the charge inside the volume whose border is the Surface $S$.Now since the electric field only depends of the distance and the considered charge in your problem is spherical symmetric you take the surface to be a sphere around the charge. That implies that the surface normal has the same direction of the electric field, such that: $\vec{E}\cdot \vec{n}=\|\vec{E}\|$ and you get the well known formula

$$\vec{E}(r)=\frac{q_I}{4\pi\epsilon_0r^2}\hat{r}.$$

To get the potential you use the definition $\vec{E}=-\nabla V$. So you get that

$$\int_a^b -\frac{dV}{dr}dr=\int_a^b E\space dr,$$ or

$$V(a)=\int_a^b E\space dr+V(b).$$

So the answer to you question: The potential at a point is only defined relatively to another point. Usually one takes the point $b$ to be in the infinite and take the potential to be zero there.

Hope this helps.

share|improve this answer
    
Thanks alot , that really helped –  Abdulrahman Hessen Mar 2 '13 at 18:45
    
@PML Two important things you don't mention: 1) the charge is on sphere, hence the field will be spherically symmetrical around the same center. 2) because of that, the integration surface is also taken to be a sphere with the same origin, which simplifies the dot product $\vec{E}\cdot \vec{n} = ||\vec{E}||$, since the two vectors are aligned everywhere on the surface. –  peterph Mar 2 '13 at 22:32
    
@peterph Thank you for the comment. I thought of that that but by some odd reason I forgot to write it. I edited the answer to include that information. Thank you –  PML Mar 3 '13 at 13:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.