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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsGauss LawMedium2 minQB

PhysicsMediuminteger

A solid sphere of radius R has a charge Q distributed in its volume with a charge density $ρ = kr^{a}$ , where K and a are constants and r is the distance from its centre. if the electric field at r= $\frac{R}{2}$ is $\frac{1}{8}$ times that at r=R, find the value of a.

Answer:

Solution:

$Q (r) = \int_{0}^{r} k r^{a} 4 π r^{2} d r = \frac{4 πk r ^{a + 3}}{a + 3}$ .

$\frac{E ( R /2 )}{E ( R )} = \frac{Q ( R /2 )}{Q ( R )} \times 4 = \frac{4}{2 ^{a + 3}} = \frac{1}{8}$ .

$2^{a + 3} = 32$ ⇒ $a = 2$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Gauss Law

⏱ 2mℹ️ Source: QB

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