Question

a solid sphere of radius R1 and volume charge density p =p0/r is enclosed by a hollow sphere of radius R2 with negative charge surface density sigma, such that th total charge in the system is zero. P0 is a positive constant and r is the distance from the centre of the sphere. The ratio R2/R1​

Answer 1

volume charge density on sphere is given , $\rho=\frac{\rho_0}{r}$

we know, volume charge density is the rate of change of charge per unit volume.

so, $\frac{dQ}{dV}=\rho$

$\implies dQ=\rho dV$

volume of sphere of radius r , V= 4/3 πr³

differentiating both sides

dV = 4πr²dr

now, charge on sphere, $dQ=\rho dV$

= $\frac{\rho_0}{r}4\pi r^2dr$

= $4\pi\rho_0r dr$

now integrating both sides,

$\int\limits^Q_0{dQ}=4\pi\rho_0\int\limits^{R_1}_0{r}\,dr$

$Q=4\pi\rho_0\frac{R_1^2}{2}$

$Q=2\pi\rho_0R_1^2$.....(1)

and surface charge density of hollow sphere of radius $R_2$ is $\sigma$

we know, $\sigma=\frac{dQ}{dA}$

or, $dQ=\sigma dA$

area of hollow sphere , A = 4πr²

differentiating both sides,

dA = 8πr dr

now, $\int\limits^Q_0{dQ}=8\pi\sigma\int\limits^{R_2}_0{r}\,dr$

$Q=4\pi\sigma R_2^2$.....(2)

from equations (1) and (2),

$2\pi\rho_0 R_1^2=4\pi\sigma R_2^2$

$\frac{R_1}{R_2}=\sqrt{\frac{2\sigma}{\rho_0}}$

hence, option (b) is correct.

Answer 2

Explanation:

Here,

charge = volume charge density × volume