2
$\begingroup$

For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$. In order to find potential of this sphere at the surface, why is my approach giving different answers?

Approach 1:

$$\rho = \frac{3Q}{4 \pi R^{3}}$$

$$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$ Potential at the surface would be $$V = \frac{q}{4 \pi \epsilon_0 x} = \frac{Q x^{2}}{4 \pi \epsilon_0 R^{3}}$$

Approach 2: $$\rho = \frac{3Q}{4 \pi R^{3}}$$ $$q = \frac{4}{3} \pi x^{3} \rho = Q \frac{x^{3}}{R^3}$$ $$E = \frac{Q x}{4 \pi \epsilon_0 R^{3}}$$ (From Gauss' Law)

Potential at the surface would be $$V = -\int{\vec{E} \cdot \vec{dx}} = -\frac{Q}{4 \pi \epsilon_0 R^{3}} \int_{0}^{x}{xdx} = -\frac{Q x^{2}}{8 \pi \epsilon_0 R^{3}}$$

Why is the answer different in both the cases?

$\endgroup$
2
$\begingroup$

Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html

$\endgroup$
2
$\begingroup$

Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(

$\endgroup$
  • 1
    $\begingroup$ You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $-\frac{3Q}{8 \pi \epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$-\frac{3Q}{8 \pi \epsilon_0 R} - ( - \frac{Q}{8 \pi \epsilon_0 R}) = \boxed{-\frac{Q}{4 \pi \epsilon_0 R}}$$ and so we are done. $\endgroup$ – PhysicsMonster_01 Apr 10 at 6:39
1
$\begingroup$

The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.
I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.

Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.

enter image description here

The area under the graph $\int E\,dx$ is related to the change in potential.

In essence what you have done is found that areas $A$ and $B$ are not the same.

PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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