A circular disc of radius R carries surface charge density \sigma \left(\right. r \left.\right) = \left(\sigma\right)_{0} \left(\right. 1 - \frac{r}{R} \left.\right), where σ0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is \left(\phi\right)_{0}. Electric flux through another spherical surface of radius \frac{R}{4} and concentric with the disc is \phi. Then the ratio \frac{\left(\phi\right)_{0}}{\phi} is________

Engineering

Physics

Magnet and Magnetic Dipole Moment

Question

A circular disc of radius R carries surface charge density $σ (r) = σ_{0} (1 - \frac{r}{R})$ , where σ₀ is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $ϕ_{0}$ . Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $ϕ$ . Then the ratio $\frac{ϕ_{0}}{ϕ}$ is_________.

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$ϕ_{0} = \frac{\int dq}{ε_{0}} = \frac{\int_{0}^{R} σ_{0} (1 - \frac{r}{R}) 2 πrdr}{ε_{0}}$

$ϕ = \frac{\int dq}{ε_{0}} = \frac{\int_{0}^{R / 4} σ_{0} (1 - \frac{r}{R}) 2 πrdr}{ε_{0}}$

$∴ \frac{ϕ_{0}}{ϕ} = = \frac{σ_{0} 2 π \int_{0}^{R} (1 - \frac{r}{R}) dr}{σ_{0} 2 π \int_{0}^{R / 4} (r - \frac{r^{2}}{R}) dr}$

$= \frac{\frac{R^{2}}{2} - \frac{R^{2}}{3}}{\frac{R^{2}}{32} - \frac{R^{2}}{3 \times 64}} = \frac{32}{5}$

= 6.40

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