A circular disc of radius carries surface charge density where is a constant and is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is . Electric flux through another

Question

A circular disc of radius $R$ carries surface charge density $σ (r) = σ_{0} (1 - \frac{r}{R}),$ where $σ_{0}$ 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 _________. [2020]

Smart Achievers · Accepted Answer

(6.40)

Let us consider a ring element of radius $σ (r) = σ_{0} (1 - \frac{r}{R}),$ and thickness dr.

Surface charge density of a disc of radius R,

$σ (r) = σ_{0} (1 - \frac{r}{R})$

Charge of disc element,

$d q = σ_{0} (1 - \frac{r}{R}) 2 π r d r$

Now, from Gauss's theorem, electric flux through a large spherical surface that encloses the charged disc completely is

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

Electric flux through another spherical surface of radius $\frac{R}{4}$

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

$∴$ $\frac{ϕ_{0}}{ϕ} = \frac{σ_{0} 2 π \int_{0}^{R} (r - \frac{r^{2}}{R}) d r}{σ_{0} 2 π \int_{0}^{R / 4} (r - \frac{r^{2}}{R}) d r}$ $= \frac{[\frac{R^{2}}{2} - \frac{R^{2}}{3}]}{[\frac{R^{2}}{32} - \frac{R^{2}}{3 \times 64}]}$

$= \frac{\frac{R^{2}}{6}}{\frac{5 R^{2}}{192}} = \frac{32}{5} = 6.40$

Solution

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