Consider a sphere of radius $R$ with charge density distributed as :
$\rho(r) =k r$, $r \leq R $
$=0$ for $r> R$.
$(a)$ Find the electric field at all points $r$.
$(b)$ Suppose the total charge on the sphere is $2e$ where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
Consider a sphere of radius $R$ with charge density distributed as :
$\rho(r) =k r$, $r \leq R $
$=0$ for $r> R$.
$(a)$ Find the electric field at all points $r$.
$(b)$ Suppose the total charge on the sphere is $2e$ where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
$(a)$ Let us consider a sphere $\mathrm{S}$ of radius $\mathrm{R}$ and two hypothetic spheres of radius $r<\mathrm{R}$ and $r>\mathrm{R}$. Electric field intensity for point $r<\mathrm{R}$,
$\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} \int \rho d \mathrm{~V}$
But, volume $\mathrm{V}=\frac{4}{3} \pi r^{3}$
$d \mathrm{~V} =\frac{4}{3} \pi \times 3 r^{2} d r$
$d \mathrm{~V} =4 \pi r^{2} d r$
$\therefore \oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{1}{\epsilon_{0}} 4 \pi k \int_{0}^{r} r^{3} d r \quad[\rho=k r]$ $\therefore(\mathrm{E}) 4 \pi r^{2}=\frac{4 \pi k}{\epsilon_{0}} \cdot \frac{r^{4}}{4}$ $\therefore \mathrm{E}=\frac{1}{4 \epsilon_{0}} \cdot k r^{2}$ As charge density is positive it means the direction of $\overrightarrow{\mathrm{E}}$ is radially outwards.
Similar Questions
Let $\rho (r)\, = \frac{Q}{{\pi {R^4}}}\,r$ be the volume 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 electric field is
Let $\rho (r)\, = \frac{Q}{{\pi {R^4}}}\,r$ be the volume 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 electric field is
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
The region between two concentric spheres ofradii '$a$' and '$b$', respectively (see figure), have volume charge density $\rho = \frac{A}{r}$ where $A$ is a constant and $r$ is the distance from the centre. At the centre of the spheres is a point charge $Q$. The value of $A$ such that the electric field in the region between the spheres will be constant, is :
The region between two concentric spheres ofradii '$a$' and '$b$', respectively (see figure), have volume charge density $\rho = \frac{A}{r}$ where $A$ is a constant and $r$ is the distance from the centre. At the centre of the spheres is a point charge $Q$. The value of $A$ such that the electric field in the region between the spheres will be constant, is :
- [JEE MAIN 2016]
Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If $\sigma _1$ is the surface charge on the inner surface and $Q_1$ net charge on it and $\sigma _2$ the surface charge on the outer surface and $Q_2$ net charge on it then
Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If $\sigma _1$ is the surface charge on the inner surface and $Q_1$ net charge on it and $\sigma _2$ the surface charge on the outer surface and $Q_2$ net charge on it then
- [JEE MAIN 2015]
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho$ (r) [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure The electric field is only along rhe radial direction.
Figure:$Image$
$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is
$(A)$ independent of a
$(B)$ directly proportional to a
$(C)$ directly proportional to $\mathrm{a}^2$
$(D)$ inversely proportional to a
$2.$ For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.
$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$
Give the answer question $1,2$ and $3.$
The nuclear charge $(\mathrm{Ze})$ is non-uniformly distributed within a nucleus of radius $R$. The charge density $\rho$ (r) [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure The electric field is only along rhe radial direction.
Figure:$Image$
$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is
$(A)$ independent of a
$(B)$ directly proportional to a
$(C)$ directly proportional to $\mathrm{a}^2$
$(D)$ inversely proportional to a
$2.$ For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.
$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$
Give the answer question $1,2$ and $3.$
- [IIT 2008]