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PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A solid sphere of radius $R_{1}$ and volume charge density $ρ = \frac{ρ_{0}}{r}$ is enclosed by a hollow sphere of radius $R_{2}$ with negative surface charge density $σ$ , such that the total charge in the system is zero, $ρ_{0}$ is a positive constant and $r$ is the distance from the centre of the sphere. The ratio $\frac{R_{2}}{R_{1}}$ is

Options:

Answer:

Solution:

For solid sphere of radius $R_{1}$

$q_{1} = \int_{0}^{R_{1}} 4 π r^{2} d r ρ$

$= \int_{0}^{R_{1}} 4 π r^{2} d r \frac{ρ_{0}}{r}$

$q_{1} = 4 π \frac{R_{1}^{2}}{2} ρ_{0}$

$q_{2} = - 4 π R_{2}^{2} σ$

$q_{1} + q_{2} = 0$

$4 π \frac{R_{1}^{2} ρ_{0}}{2} - 4 π R_{2}^{2} σ = 0$

${(\frac{R_{1}}{R_{2}})}^{2} = \frac{2 σ}{ρ_{0}}$

$\frac{R_{2}}{R_{1}} = \sqrt{\frac{ρ_{0}}{2 σ}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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