Question

Total charge of the sphere will be?

Answer 1

Answer :

Chrge density of sphere = $\sf{\dfrac{\rho_o}{r}}$

where r is the distance from centre of sphere$.$

Radius of sphere = R

We have to find total charge of sphere.

____________________________

◈ Let's consider a sphere of radius r (thickness dr) inside the given sphere of radius R.

Let charge enclosed by this sphere be dq and volume of this sphre be dV.

Volume of sphere is given by

⇒ V = 4/3 πr³

⇒ dV = 4πr² (dr)

Charge enclosed by sphere in terms of density is given by

⇒ dq = ρ × dV

Net charge enclosed by sphere of radius R is given by

$\displaystyle:\implies\sf\:q=\int{dq}=\int{\dfrac{\rho_o}{r}\times 4\pi r^2\:(dr)}$

$\displaystyle:\implies\sf\:q=4\rho_o \pi\int{r\:(dr)}$

$:\implies\sf\:q=4\rho_o\pi\times \dfrac{r^2}{2}$

Taking limit 0 ➝ R

$:\implies\sf\:q=4\rho_o\pi\times \dfrac{R^2}{2}$

$:\implies\boxed{\bf{\red{q=2\rho_o\pi R^2}}}$

Hence, (2) is the correct answer!

Answer 2

ANSWER.

option [ 2 ] is correct answer.

EXPLANATION.

$\sf : \implies \: charge \: density \: of \: a \: sphere \: of \: radius \:R \: is \: \frac{ \rho}{r}$

$\sf : \implies as \: we \: know \: that \\ \\ \sf : \implies \: dq \: = \rho \: dv \\ \\ \sf : \implies \: dv = (4\pi \: {r}^{2}dr) \\ \\ \sf : \implies \: as \: we \: know \: that \\ \\ \sf : \implies \: dq = \rho(4\pi \: {r}^{2}dr) \\ \\ \sf : \implies \: \rho \: = \frac{ \rho_{o} }{r} \: \: \: \: \: (given)$

$\sf : \implies \: dq = \dfrac{ \rho_{o} }{r} (4\pi \: {r}^{2} dr) \\ \\ \sf : \implies \: \int \limits_{0} {}^{Q} \: \: (dq) = 4\pi \: \rho_{o} \int \limits_{0} {}^{R } \: \: (dr ) \\ \\ \sf : \implies \: q \: = 4\pi \: \rho_{o} \:( \frac{ {r}^{2} }{2} ) \\ \\ \sf : \implies \: 2\pi \: \rho_{o} \: R {}^{2}$