Question

When a point charge \[+q\] is embedded in a medium of dielectric constant \[K\], the effective permittivity of the medium is:

Options

• \[\varepsilon = \varepsilon_0 + K\]

• \[\varepsilon = \varepsilon_0\] regardless of \[K\]

• \[\varepsilon = \frac{\varepsilon_0}{K}\]

• \[\varepsilon = K\varepsilon_0\]

Answer 1

In a medium of dielectric constant \[K\), the effective permittivity is \[\varepsilon = K\varepsilon_0\). This reduces the force on a unit positive test charge to \[F=\frac{1}{4\pi\varepsilon_0K}\frac{q}{x^2}\), and hence the potential to \[V(r)=\frac{q}{4\pi\varepsilon_0Kr}\).