The electrostatic potential inside a charged spherical ball is given by $$\phi = a{r^2} + b$$ where $$r$$ is the distance from the center and $$a,b$$ are constants. Then the charge density inside the ball is:

Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d\left( {d < < 1} \right)$$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $$v$$. Then as a function of distance $$x$$ between them,

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $${30^ \circ }$$ with each other. When suspended in a liquid of density $$0.8g$$ $$c{m^{ - 3}},$$ the angle remains the same. If density of the material of the sphere is $$1.6$$ $$g$$ $$c{m^{ - 3}},$$ the dielectric constant of the liquid is

Let there be a spherically symmetric charge distribution with charge density varying as $$\rho \left( r \right) = {\rho _0}\left( {{5 \over 4} - {r \over R}} \right)$$ upto $$r=R,$$ and $$\rho \left( r \right) = 0$$ for $$r>R,$$ where $$r$$ is the distance from the erigin. The electric field at a distance $$r\left( {r < R} \right)$$ from the origin is given by