Question

The total work done in bringing a unit positive test charge from infinity to a distance \[r\] from a point charge \[+q\] in vacuum is:

पर्याय

• \[W=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r^2}-\frac{1}{\infty^2}\right)=\frac{q}{4\pi\varepsilon_0 r^2}\]

• \[W=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r}-\frac{1}{\infty}\right)=\frac{q}{4\pi\varepsilon_0 r}\]

• \[W=\frac{q}{4\pi\varepsilon_0 r}\left(1-\frac{r}{\infty}\right)=0\]

• \[W=-\frac{q}{4\pi\varepsilon_0 r}\]

Answer 1

Integrating \[dW=-\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}dx\] from \[\infty\] to \[r\] gives \[W=\frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r}-\frac{1}{\infty}\right)=\frac{q}{4\pi\varepsilon_0 r}\], since \[\frac{1}{\infty}=0\]. For a unit test charge, \[V(r)=W\].