Question

For a volume charge distribution, what is the integral form of the electric field?

Options

• \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\sum\frac{\rho dV}{r^{\prime2}}\]

• \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho dV}{r^{\prime2}}\hat{r}^{\prime}\]

• \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int\frac{\lambda dl}{r^{\prime2}}\hat{r}^{\prime}\]

• \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int\frac{\sigma dS}{r^{\prime2}}\hat{r}^{\prime}\]

Answer 1

For a volume charge distribution, the electric field is obtained by integrating the field contributions from all infinitesimal volume elements. The integral form is \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int\frac{\rho dV}{r^{\prime2}}\hat{r}^{\prime}\], where ρ is the volume charge density and dV is the volume element.