Explain by graph how the electric field by thin spherical shell depends on the distance of point from centre.

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A conducting sphere of radius $10\, cm$ has unknown charge. If the electric field at a distance $20\, cm$ from the centre of the sphere is $1.2 \times 10^3\, N\, C^{-1}$ and points radially inwards. The net charge on the sphere is

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]

Obtain the expression of electric field by ......

$(i)$ infinite size and with uniform charge distribution.

$(ii)$ thin spherical shell with uniform charge distribution at a point outside it.

$(iii)$ thin spherical shell with uniform charge distribution at a point inside it.

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.

A uniform rod $AB$ of mass $m$ and length $l$ is hinged at its mid point $C$ . The left half $(AC)$ of the rod has linear charge density $-\lambda $ and the right half $(CB)$ has $+\lambda $  where $\lambda $ is constant . A large non conducting sheet of unirorm surface charge density $\sigma $ is also .present near the rod. Initially the rod is kept perpendicular to the sheet. The end $A$ of the rod is initially at a distance $d$ . Now the rod is rotated by a small angle in the plane of the paper and released. The time period of small angular oscillations is