A circular disc of radius R caries surface charge density $σ (r) = σ_{0} (1 - \frac{r^{2}}{R^{2}}),$ where $σ_{0}$ is a constant and r is the distance from the centre of the disc. Assume that disc is in xz – plane and its axis is y–axis. We have an imaginary hollow sphere of radius R and centre at $(0, \frac{3}{5} R, 0) .$ What is the flux of electric field through this imaginary sphere?

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$\frac{136}{625} (\frac{σ_{0} π R^{2}}{ε_{0}})$

$\frac{544}{625} (\frac{σ_{0} π R^{2}}{ε_{0}})$

$\frac{408}{625} (\frac{σ_{0} π R^{2}}{ε_{0}})$

$\frac{272}{625} (\frac{σ_{0} π R^{2}}{ε_{0}})$

answer is B.

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Detailed Solution

Question Image
Charge on circular part
$q = \int σ (r) . 2 π r . d r$
= $\int_{0}^{(4 / 5) R} 2 π σ_{0} (r - \frac{r^{3}}{R^{2}}) d r = \frac{272}{625} (σ_{0} π R^{2})$
$ϕ = \frac{q}{ε_{0}} = \frac{272}{625} (\frac{σ_{0} π R^{2}}{ε_{0}})$

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