An infinitely long solid cylinder of radius R | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The electric field at point $P$ is the vector sum of the electric field due to the solid cylinder (without the cavity) and the electric field due

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The electric field at point $P$ is the vector sum of the electric field due to the solid cylinder (without the cavity) and the electric field due to a sphere of radius $R/2$ with charge density $-\rho$.$1$. Electric field due to the solid cylinder at distance $r = 2R$:Using Gauss's Law,$E_1 = \frac{\lambda}{2 \pi \varepsilon_0 r}$,where $\lambda = \rho \pi R^2$.$E_1 = \frac{\rho \pi R^2}{2 \pi \varepsilon_0 (2R)} = \frac{\rho R}{4 \varepsilon_0}$.$2$. Electric field due to the spherical cavity (treated as a sphere of charge density $-\rho$):The charge of the sphere is $q = -\rho \cdot \frac{4}{3} \pi (R/2)^3 = -\rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8} = -\frac{\rho \pi R^3}{6}$.The electric field at distance $2R$ from the center is $E_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q|}{(2R)^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \pi R^3 / 6}{4R^2} = \frac{\rho R}{96 \varepsilon_0}$.$3$. Net electric field $E = E_1 - E_2$:$E = \frac{\rho R}{4 \varepsilon_0} - \frac{\rho R}{96 \varepsilon_0} = \frac{\rho R}{\varepsilon_0} \left( \frac{24 - 1}{96} \right) = \frac{23 \rho R}{96 \varepsilon_0}$.Given $E = \frac{23 \rho R}{16 k \varepsilon_0}$,we have $\frac{23 \rho R}{96 \varepsilon_0} = \frac{23 \rho R}{16 k \varepsilon_0}$.Thus,$96 = 16k \Rightarrow k = 6$.

Similar Questions

Three infinitely long charged non-conducting sheets are placed as shown in the figure. The electric field at point $P$ is ($\sigma$ - charge density,$\epsilon_0$ - permittivity of free space).

The electrostatic field of a long uniformly charged wire varies with distance $r$ according to which relation?

Two parallel large thin metal sheets have equal surface charge densities $(\sigma = 26.4 \times 10^{-12} \, C/m^2)$ of opposite signs. The electric field between these sheets is

$A$ hollow insulated conducting sphere is given a positive charge of $10\,\mu C$. What will be the electric field at the centre of the sphere if its radius is $2\,m$?

$A$ non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre:

Vedclass Test Series

Exam Paper Generator

Online Exam Module