1
$\begingroup$

Consider a positive charge $Q$ at the origin.What is the work done in bringing a unit positive charge from infinity to zero. We know that the work done in bringing the charge from infinity to $r$ is the electrostatic potential $$V(r)=\dfrac Q{4\pi\epsilon_0 r}$$

But what is the work done to bring it from infinity to $0$, i.e,when $r=0$? Can we bring it initially from infinity to $r$ and then from $r$ to $0$? Then what is the work done in bringing a unit charge from $r$ to zero?

$\endgroup$
  • 1
    $\begingroup$ The Greek letter is "epsilon", not "absolon" $\endgroup$ – Aaron Stevens Aug 1 at 3:40
3
$\begingroup$

It would take an infinite amount of work to bring the charge to $r=0$ where $Q$ is, regardless of whether you start infinitely far away or finitely far away. The repulsive force between the two charges gets infinitely strong as they get infinitely close, and you have to work against it.

$\endgroup$
0
$\begingroup$

Work done bringing charge from infinity to the origin doesn’t imply that work done is zero, because the origin is itself is a part of coordinate system but will it requires infinite amount of work that is need to done for bringing the charge from infinity to the origin where $+q$ charge is already there, we can easily understand by the given equation

The electrostatic potential energy, $U_E$, of one point charge $q$ at position $r$ in the presence of a point charge $Q$, taking an infinite separation between the charges as the reference position, is:

$$U_{E}(r)=k_{e}{\frac {qQ}{r}},$$

where $k_{e}={\frac {1}{4\pi \varepsilon _{0}}}$ is Coulomb's constant, $r$ is the distance between the point charges $q$ and $Q$, and $q$ and $Q$ are the charges (not the absolute values of the charges — i.e., an electron would have a negative value of charge when placed in the formula). The following outline of proof states the derivation from the definition of electric potential energy and Coulomb's law to this formula.

Outline of proof

definition of the position vector r and the displacement vector s, it follows that r and s are also radially directed from Q. So, E and ds must be parallel:

{\displaystyle \mathbf {E} \cdot \mathrm {d} \mathbf {s} =|\mathbf {E} |\cdot |\mathrm {d} \mathbf {s} |\cos(0)=E\mathrm {d} s} {\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {s}}=|{\mathbf {E}}|\cdot |{\mathrm {d}}{\mathbf {s}}|\cos(0)=E{\mathrm {d}}s Using Coulomb's law, the electric field is given by

{\displaystyle |\mathbf {E} |=E={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{s^{2}}}} |{\mathbf {E}}|=E={\frac {1}{4\pi \varepsilon _{0}}}{\frac {Q}{s^{2}}} and the integral can be easily evaluated:

{\displaystyle U_{E}(r)=-\int _{\infty }^{r}q\mathbf {E} \cdot \mathrm {d} \mathbf {s} =-\int _{\infty }^{r}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{s^{2}}}{\rm {d}}s={\frac {1}{4\pi \varepsilon {0}}}{\frac {qQ}{r}}=k{e}{\frac {qQ}{r}}} U_{E}(r)=-\int _{\infty }^{r}q{\mathbf {E}}\cdot {\mathrm {d}}{\mathbf {s}}=-\int _{\infty }^{r}{\frac {1}{4\pi \varepsilon _{0}}}{\frac {qQ}{s^{2}}}{{\rm {d}}}s={\frac {1}{4\pi \varepsilon {0}}}{\frac {qQ}{r}}=k{e}{\frac {qQ}{r}} One point charge q in the presence of n point charges Qi
Edit

Electrostatic potential energy of q due to Q1 and Q2 charge system: {\displaystyle U_{E}=q{\frac {1}{4\pi \varepsilon {0}}}\left({\frac {Q{1}}{r_{1}}}+{\frac {Q_{2}}{r_{2}}}\right)} U_{E}=q{\frac {1}{4\pi \varepsilon {0}}}\left({\frac {Q{1}}{r_{1}}}+{\frac {Q_{2}}{r_{2}}}\right) The electrostatic potential energy, UE, of one point charge q in the presence of n point charges Qi, taking an infinite separation between the charges as the reference position, is:

{\displaystyle U_{E}(r)=k_{e}q\sum {i=1}^{n}{\frac {Q{i}}{r_{i}}}} U_{E}(r)=k_{e}q\sum {{i=1}}^{n}{\frac {Q{i}}{r_{i}}},

$\endgroup$
  • 1
    $\begingroup$ I’ve fixed the MathJax in the first few paragraphs so that it actually displays math now. You can fix the rest by looking at what I did. $\endgroup$ – G. Smith Aug 1 at 3:33
0
$\begingroup$

I'll again repeat the usual saying that the absolute value of potential energy doesn't matter. What matters is the relative difference.

Having said that we can define the potential energy of a charge at some point relative to some other point. You can think of potential energy as the energy of a system, and not that of a single charge. When we say the charge is at infinity we mean that there are no other charges in the vicinity of the test charge and hence no force is felt by our little test charge. Hence its potential energy is zero. But you are free to define any point of your interest where potential energy is zero.

Remember, once you've defined your zero PE point, the PE's at all other points are obvious.

But what is the work done to bring it from infinity to $0$, i.e, when $r=0$?

In your case, the potential energy of both the $Q$ and the test charge will increase by the same amount. Putting $r=0$ in your formula will give infinite potential which means you'd have to spend infinite amount of energy to to reduce the distance $r$ between the charges to zero which doesn't make sense.

Can we bring it initially from infinity to $r$ and then from $r$ to $0$?

Remember potential energy is a state function. It really doesn't matter which way you do it. Only the beginning and the end points matter. The change in PE will be the same in both the cases whether you bring it directly from $\infty$ to $0$ or $\infty$ to $r$ and then from $r$ to $0$. But as I said before potential at $r=0$ doesn't make sense.

$\endgroup$

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.