Let P(r) = frac{Q}{pi R^4} r be the charge de | vedclass

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

Question

(C) According to Gauss's Law,the electric field $E$ at a distance $r_1$ from the center is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilo

Vedclass · Accepted Answer

$\frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$

Vedclass · Answer

(C) According to Gauss's Law,the electric field $E$ at a distance $r_1$ from the center is given by $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.For a spherical surface of radius $r_1$,this becomes $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \rho(r) (4\pi r^2) dr$.Substituting $\rho(r) = \frac{Q}{\pi R^4} r$,we get $E(4\pi r_1^2) = \frac{1}{\varepsilon_0} \int_0^{r_1} \left( \frac{Q}{\pi R^4} r \right) (4\pi r^2) dr$.$E(4\pi r_1^2) = \frac{4\pi Q}{\pi R^4 \varepsilon_0} \int_0^{r_1} r^3 dr$.$E(4\pi r_1^2) = \frac{4 Q}{R^4 \varepsilon_0} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4 \varepsilon_0}$.Solving for $E$,we get $E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

List-$I$	List-$II$
$(A)$ Electric field inside (distance $r < R$ from center) of a uniformly charged spherical shell with surface charge density $\sigma$ and radius $R$.	$(I)$ $\sigma / \varepsilon_0$
$(B)$ Electric field at distance $r$ from a uniformly charged infinite plane sheet with surface charge density $\sigma$.	$(II)$ $\sigma / 2 \varepsilon_0$
$(C)$ Electric field outside (distance $r > R$ from center) of a uniformly charged spherical shell with surface charge density $\sigma$ and radius $R$.	$(III)$ $0$
$(D)$ Electric field between $2$ oppositely charged infinite plane parallel sheets with uniform surface charge density $\sigma$.	$(IV)$ $\frac{\sigma R^2}{\varepsilon_0 r^2}$

Let $P(r) = \frac{Q}{\pi R^4} r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $P$ inside the sphere at distance $r_1$ from the centre of the sphere,the magnitude of the electric field is

Similar Questions

$A$ very long charged solid cylinder of radius 'a' contains a uniform charge density $\rho$. The dielectric constant of the material of the cylinder is $k$. What will be the magnitude of the electric field at a radial distance '$x$' $(x < a)$ from the axis of the cylinder?

$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is:

The electric field intensity on the surface of a solid charged sphere of radius $r$ and volume charge density $\rho$ is ( $\epsilon_0=$ permittivity of free space).

An infinitely long thin straight wire has a uniform linear charge density of $\frac{1}{3} \, C \cdot m^{-1}$. The magnitude of the electric field intensity at a point $18 \, cm$ away is (given $\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 \cdot N^{-1} \cdot m^{-2}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module