Considering a strip of width dx at a distance x from centre,
$\text{di}\frac{\text{dp}}{\text{dt}}=\frac{\text{q}4\pi\text{x}^2\text{dx}}{\Big(\frac{4}{3}\Big)\pi\text{R}^3\text{}}4\pi\text{x}^2\text{dx}$
$\text{di}=\frac{\text{dq}}{\text{dt}}=\frac{\text{q}4\pi\text{x}^2\text{dx}}{\Big(\frac{4}{3}\Big)\pi\text{R}^3\text{t}}=\frac{3\text{q}\text{x}^2\text{dx}\omega}{\text{R}^32\pi}$
$\text{d}\mu=\text{di}\times\text{A}=\frac{3\text{q}\text{x}^2\text{d}\text{x}\omega}{\text{R}^32\pi}\times4\pi\text{x}^{2\ =\ \frac{6\text{q}\omega}{\text{R}^3}\text{x}^4\text{dx}}$
$\mu=\int\limits_0^{\mu}\text{d}\mu\int\limits_0^{\text{R}}\frac{6\text{q}\omega}{\text{R}^3}\text{x}^4\text{dx}=\frac{6\text{q}\omega}{\text{R}^3}\Big[\frac{\text{x}^5}{5}\Big]_0$
$\text{R}=\frac{\text{6}\text{q}\omega\text{R}^5}{\text{R}^3}\frac{\text{R}^5}{5}=\frac{6}{5}\text{q}\omega\text{R}^2$
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AC = CO = D, S1C = S2C = d < < D
A small transparent slab containing material of μ = 1.5 is placed along AS2 (Fig). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab.
