Consider a long uniformly charged cylinder having constant volume charge density ' $\lambda$ ' and radius '…

Consider a long uniformly charged cylinder having constant volume charge density ' $\lambda$ ' and radius ' $R$ '. A Gaussian surface is in the form of a cylinder of radius ' $r$ ' such that vertical axis of both the cylinders coincide. For a point inside the cylinder $(r \lt R)$, electric field is directly proportional to
  1. $\mathrm{r}^{-1}$
  2. r
  3. $\mathrm{r}^2$
  4. $\mathrm{r}^{-2}$

Solution

$\begin{array}{ll} & \lambda=\frac{\mathrm{q}}{\mathrm{V}} \quad \therefore \mathrm{q}=\lambda \mathrm{V} \\ \therefore \quad & \mathrm{q}=\lambda \pi \mathrm{r}^2 \mathrm{~L} \\ & \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{\mathrm{q}}{\varepsilon_0} \\ \therefore \quad & \mathrm{E} \int \mathrm{dA}=\frac{\lambda \pi \mathrm{r}^2 \mathrm{~L}}{\varepsilon_0} \\ \therefore \quad & \mathrm{E}(2 \pi \mathrm{rL})=\frac{\lambda \pi \mathrm{r}^2 \mathrm{~L}}{\varepsilon_0} \\ \therefore \quad & \mathrm{E}=\frac{\lambda \mathrm{r}}{2 \varepsilon_0} \\ \therefore \quad & \mathrm{E} \propto \mathrm{r}\end{array}$

Asked in: MHT CET 2024 (15 May Shift 1)

Practice more Electrostatics questions on Aicharya