According to my book, 'The potential at a point is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.' But I wonder how it is possible. As the charge is being brought from infinity, the work done = force * infinity, thus, the work done would be infinity indeed. Please help me out.
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4 Answers
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Work isn’t $Fd$ when the force changes with position. It’s an integral, and the integral is finite even over an infinite distance because the force goes to zero sufficiently rapidly at large distances.
The integral is a standard homework problem so I am not going to write it or evaluate it.
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Loosely speaking, at great distance the force required gets infinitely tiny, so the two infinities largely cancel out.
answered Oct 11, 2019 at 18:29
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The force between two point charges is an example of an inverse square law where the force is proportional to the reciprocal of the separation squared, $F\propto \frac {1}{r^2}$, as illustrated below.
As the separation increases so does the force decrease.
The area under a force against separation graph is the work done and as long as the area under the graph from $r=R$ to $r\, (=) \, \infty$ is finite, which it is, there is no problem.
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Force is dependent of distance $r$, so thing move like this
$$\newcommand{\newln}{\\&\quad\quad{}} \begin{align}&\int^{r_b}_{r_a}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_a-U_b) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}}=-(U_r-U_\infty) \newln \Rightarrow \int^{r}_{\infty}\mathbf{\vec{F}}\cdot d\mathbf{\vec{r}} =-U_r ~~~~~~~ [U_\infty = 0]\newln \Rightarrow \int^{r}_{\infty}k\cdot\frac{q.q_o}{r^2}d\mathrm{\mathbf{r}}=-U_r ~~~~~~~ [\textrm{Coulomb's Law}]\newln \Rightarrow kq\cdot q_o\int^{r}_{\infty}\frac{1}{r^2}d\mathbf{r}=-U_r\newln \Rightarrow kq\cdot q_o\left[\frac{-1}{r} \right]^r_\infty=-Ur\newln\Rightarrow \frac{-kq.q_o}{r}=-U_r\newln \Rightarrow U_r=\frac{kq.q_o}{r} \end{align} $$, And potential energy is $(-) work done by conservative force.
