TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, Line charge density of rod, $\lambda = \frac{q}{l}$ . . . (i) where, $q$ is total charge of rod and $L$ is the total length of rod.Now, $dq = \lambda dx$ where, $dq$ is elemental charge at distance $x$ from origin and $dx$ is elemental length. $\because \; dE_y = \frac{k dq}{\sqrt{x^2+y^2}} \cos \theta$ where, $dE$ is elemental electric field. $\therefore \; dE = k \frac{\lambda dx}{(\sqrt{x^2+y^2})^2} \cdot \frac{x}{\sqrt{x^2+y^2}}$ $\Rightarrow \; dE = k\lambda \cdot \frac{x dx}{(x^2+y^2)^{3/2}}$. . . (ii) Let, $x^2 + y^2 = r^2$ On differentiating both sides w.r.t $x$, we get $2x + 0 = 2r \frac{dr}{dx}$ $\Rightarrow \; x dx = r dr$ Substituting the value Eq. (ii), we get $dE = k\lambda \frac{r dr}{(r^2)^{3/2}}$ On integrating both sides, we get $\int\limits_{0}^{E} dE = \int\limits_{0}^{r} k\lambda \frac{r dr}{r^3} = k\lambda \int\limits_{0}^{r} r^{-2} dr = k\lambda \left( \frac{r^{-1}}{-1} \right)_{0}^{r} = -\frac{k\lambda}{r}$ $= -\frac{1}{4\pi\varepsilon_0} \frac{\lambda}{\sqrt{x^2+y^2}} = \frac{1}{4\pi\varepsilon_0} \frac{\lambda}{\sqrt{\left(\frac{L}{2}\right)^2 + \left(\frac{L}{2}\right)^2}}$ $= \frac{1}{4\pi\varepsilon_0} \frac{\lambda}{\sqrt{\frac{L^2}{4}+\frac{L^2}{4}}} = \frac{1}{4\pi\varepsilon_0} \frac{\lambda}{\sqrt{\frac{L^2}{2}}} = \frac{1}{2\sqrt{2}\pi\varepsilon_0} \frac{\lambda}{L}$