Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{ I }, E_{ II }$ and $E_{III}$ are
$\vec{E}_{ I }=\frac{2 \sigma}{\epsilon_0} \hat{n}, \vec{E}_{ II }=0, \vec{E}_{ III }=\frac{2 \sigma}{\epsilon_0} \hat{n}$
$\vec{E}_{ I }=0, \vec{E}_{ II }=\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{ III }=0$
$\vec{E}_{ I }=\frac{\sigma}{2 \epsilon_0} \hat{n}, \vec{E}_{\text {II }}=0, \vec{E}_{ III }=\frac{\sigma}{2 \epsilon_0} \hat{n}$
$\vec{E}_{ I }=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{\text {II }}=0, \vec{E}_{\text {III }}=\frac{\sigma}{\epsilon_0} \hat{n}$
The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero ?
Charges $Q, 2Q$ and $4Q$ are uniformly distributed in three dielectric solid spheres $1,2$ and $3$ of radii $R/2, R$ and $2 R$ respectively, as shown in figure. If magnitudes of the electric fields at point $P$ at a distance $R$ from the centre of spheres $1,2$ and $3$ are $E_1 E_2$ and $E_3$ respectively, then
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 $3.$
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
Two parallel infinite line charges with linear charge densities $+\lambda\; \mathrm{C} / \mathrm{m}$ and $-\lambda\; \mathrm{C} / \mathrm{m}$ are placed at a distance of $2 \mathrm{R}$ in free space. What is the electric field mid-way between the two line charges?