Question

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Answer 1

Step-by-step explanation:

Given :

Radius of solid sphere = R₁

Volume charge density ρ = ρ｡/r

Radius of hollow sphere = R₂

Surface charge density = -σ

Total charge in the system is zero.

To Find :

The ratio of R₂/R₁.

Solution :

First of all we have to find individual charge on each sphere.

A] Charge on Hollow Sphere :

$\sf:\implies\:Surface\:charge\:density=\dfrac{Charge}{Area}$

$\sf:\implies\:-\sigma=\dfrac{Q_2}{4\pi R_2^2}$

$\bf:\implies\:Q_2=-\sigma\cdot 4\pi R_2^2$

B] Charge on Solid Sphere :

$\sf:\implies\:Volume\:charge\:density=\dfrac{Charge}{Volume}</p><p>\sf:\implies\:\rho=\dfrac{dQ_1}{dV}</p><p>$

$\sf:\implies\:dQ_1=\rho\:dV$

V = 4/3 πr³

dV = 4/3 π (3r²) dr

dV = 4πr² dr

$\displaystyle\sf:\implies\:Q_1=\int\limits_0^{R_1}\dfrac{\rho_o}{r}\:4\pi r^2\:dr$

$\displaystyle\sf:\implies\:Q_1=4\pi\rho_o\int\limits_0^{R_1}r\:dr</p><p>$

$\sf:\implies\:Q_1=4\pi\rho_o\cdot\dfrac{R_1^2}{2}$

$\bf:\implies\:Q_1=2\pi\rho_oR_1^2$

ATQ, Q₁ + Q₂ = 0

$</p><p>\sf:\implies\:2\pi\rho_o R_1^2+(-\sigma 4\pi R_2^2)=0$

$\sf:\implies\:2\pi(\rho_o R_1^2-2\sigma R_2^2)=0$

$\sf:\implies\:\rho_oR_1^2=2\sigma R_2^2$

$\sf:\implies\:\dfrac{R_2^2}{R_1^2}=\dfrac{\rho_o}{2\sigma}$

$:\implies\:\underline{\boxed{\bf{\orange{\dfrac{R_2}{R_1}=\sqrt{\dfrac{\rho_o}{2\sigma}}}}}}$