Our teacher said

When charge distribution is to infinity ,the potential cannot be assumed at infinity so it should be assumed zero at some arbitrary point

But why can't we take it at infinity ? Even if we see second statement can't arbitrary point be at infinity? I don't get it


1 Answer 1


As an example, consider an infinite line of charge with current density $\lambda$. Using Gauss's law you can find the electric field to be $$\vec{E} = \frac{\lambda}{2\pi\epsilon_0 \rho}\hat{\rho}.$$ To find the potential at some distance $r$, we integrate from the point at which we define the zero of the potential to r. Let's define this point to be at infinity. So we have $$V(r) = \frac{\lambda}{2\pi\epsilon_0}\int_{r}^{\infty} \frac{d\rho}{\rho}$$ $$V(r) = \frac{\lambda}{2\pi\epsilon_0}\ln{\rho}|^{\infty}_r.$$ You can immediately see that the potential evaluated at these bounds blows up. This is because the charge distribution goes out to infinity. So it appears that you can define the zero to be at infinity but you will run into trouble with infinities.

However, since we can always redefine the potential up to some arbitrary constant, we can define the zero of the potential to be at some point $R$. From this we have $$V(r) = \frac{\lambda}{2\pi\epsilon_0}\ln{\rho}|^R_r.$$ This is well-behaved for finite $R\ne 0$.


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.