A disc of radius R is charged on its surface | vedclass

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(A) The total charge $q_{	ext{total}}$ on the disc is calculated by integrating the surface charge density over the area of the disc.Consider a small

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$\frac{2 \pi \sigma_0 R^5}{5}$

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(A) The total charge $q_{	ext{total}}$ on the disc is calculated by integrating the surface charge density over the area of the disc.Consider a small elemental ring of radius $r$ and width $dr$. The area of this ring is $ds = 2 \pi r dr$.The total charge is given by:$q_{	ext{total}} = \int_0^R \sigma ds$Substituting $\sigma = \sigma_0 r^3$ and $ds = 2 \pi r dr$:$q_{	ext{total}} = \int_0^R (\sigma_0 r^3)(2 \pi r dr)$$q_{	ext{total}} = 2 \pi \sigma_0 \int_0^R r^4 dr$Performing the integration:$q_{	ext{total}} = 2 \pi \sigma_0 \left[ \frac{r^5}{5} ight]_0^R$$q_{	ext{total}} = \frac{2 \pi \sigma_0 R^5}{5}$

$A$ disc of radius $R$ is charged on its surface with surface charge density $\sigma = \sigma_0 r^3$. Here $\sigma_0$ is a constant and $r$ is the distance from its centre. The total charge on the disc is:

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