A uniformly charged conducting sphere of diameter 14 cm has surface charge density of $40 \mu \mathrm{Cm}^{-2}$. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space $=8.85 \times 10^{-12}$ SI unit)

The electrostatic potential inside a charged spherical ball is given by $\mathrm{V}=\mathrm{ar}^2+\mathrm{b}$ where ' r ' is the distance from its centre and ' $a$ ' and ' $b$ ' are constants. The volume charge density of the ball is [ $\varepsilon_0=$ permittivity of free space $]$

A charge $$17.7 \times 10^{-4} \mathrm{C}$$ is distributed uniformly over a large sheet of area $$200 \mathrm{~m}^2$$. The electric field intensity at a distance $$20 \mathrm{~cm}$$ from it in air will be $$\left[\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\right]$$

If $$\mathrm{E}_{\mathrm{a}}$$ and $$\mathrm{E}_{\mathrm{q}}$$ represent the electric field intensity due to a short dipole at a point on its axial line and on the equatorial line at the same distance '$$r$$' from the centre of the dipole, then