Question

(a) Use Gauss's theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density $\sigma$ (b) An infinitely large thin plane sheet has a uniform surface charge density $+\sigma$ . Obtain the expression for the amount of work done in bringing a point charge q from infinity. to a point, distant r, in front of the charged plane sheet.

Answer 1

(a)

Let electric charge be uniformly distributed over the surface of a thin non- conducting infinite sheet.
Let the surface charge density be $\sigma$. According to Gauss theorem : ${\int }_1\overrightarrow{E}.\overrightarrow{ds}+{\int }_2\overrightarrow{E}.\overrightarrow{ds}+{\int }_3\overrightarrow{E}.\overrightarrow{ds}=\frac{q}{{ϵ}_0}{\int }_{1}Edscos{0}^{\circ }+{\int }_{2}Edscos{0}^{\circ }+{\int }_{3}Edscos{90}^{\circ }=\frac{q}{{ϵ}_0}{\int }_{1}Eds+{\int }_{2}Eds=\frac{q}{{ϵ}_0}2\int Eds=\frac{q}{{ϵ}_0}2E.S=\frac{\sigma S}{{ϵ}_0}E=\frac{\sigma }{2{ϵ}_0}$ (b) The electric field due to a uniformly charged infinitely large thin sheet with surface charge density $\sigma$ is, $E=\frac{\sigma }{2{ϵ}_0}...\left(i\right)$ The amount of work done in bringing a point charge q from infinity to a point, at a distance r is given by $W={\int }_{\infty }^{r}F.dr$ Where F is the force experienced by charge q. Now, $W=q{\int }_{\infty }^r-E.dr[\because F=-Eq]W=-q\frac{\sigma }{2{ϵ}_0}{\int }_{\infty }^{r}dr\left[from\right(i\left)\right]W=q\frac{\sigma }{2{ϵ}_0}[\infty -r]=\infty$