Question

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

पर्याय

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

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

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

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

Answer 1

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