Question

In the derivation of the potential in a medium of dielectric constant \[K\), the total work done in bringing a unit positive test charge from infinity to distance \[r\) is:

Options

• \[W=\int_{\infty}^{r}dW=-\frac{q}{4\pi\varepsilon_{0}K}\int_{\infty}^{r}x^{-2}dx=\frac{q}{4\pi\varepsilon_{0}Kr}\]

• \[W=\frac{q}{4\pi\varepsilon_{0}}\int_{\infty}^{r}x^{-2}dx=\frac{q}{4\pi\varepsilon_{0}r}\] with no \[K\] dependence

• \[W=\frac{q}{4\pi\varepsilon_{0}Kr^2}\]

• \[W=-\frac{q}{4\pi\varepsilon_{0}Kr}\]

Answer 1

The force in the medium is \[F=\frac{1}{4\pi\varepsilon_0K}\frac{q}{x^2}\], giving \[dW=-F\,dx\]. Integrating from \[\infty\] to \[r\] yields \[W=\frac{q}{4\pi\varepsilon_{0}Kr}\], which is the potential in the medium since a unit test charge is used.