Consider a long uniformly charged cylinder having constant volume charge density $\rho$ and radius $R$. $A$ Gaussian surface is in the form of a cylinder of radius $r$ such that the vertical axis of both cylinders coincide. For a point inside the cylinder $(r < R)$,the electric field is directly proportional to

An infinite line charge produces a field of $9 \times 10^4 \ NC^{-1}$ at a distance of $2 \ cm$. Its linear charge density is

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 $\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 \cdot N^{-1} \cdot m^{-2}$):

The electric field components in the figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \; N/C \cdot m^{1/2}.$ Calculate
$(a)$ the flux through the cube,and
$(b)$ the charge within the cube. Assume that $a=0.1 \; m$.

$A$ uniform rod $AB$ of mass $m$ and length $l$ is hinged at its midpoint $C$. 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 uniform 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 $\theta$ in the plane of the paper and released. The time period of small angular oscillations is:

$A$ cube is placed inside an electric field,$\overrightarrow{E} = 150 y^2 \hat{j}$. The side of the cube is $0.5 \, m$ and it is placed in the field as shown in the figure. The charge inside the cube is $..... \times 10^{-11} \, C$.