Question

Calculate the electrostatic field intensity on the axis of an circular disc having charge q.

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Answer 1

To find electric field at point P (shown in the fig. attached) due to the disc let us consider an elemental ring of radius y and width dy. If dq be the charge on this infinitesimal element, then

dy = σ(2πydy) [Area of element is dA = 2πydy]

As we know that electric field strength due to a ring of radius R, charge Q, at a distance x from its centre on its axis can be given as

$\rm E_{axis} = \dfrac{1}{4 \pi \epsilon _0} \dfrac{Qx}{{(x^2 + R^2)}^{3/2}}$

So, due to the infinitesimal elemental ring the electric field strength dE at point P is

$\rm dE = \dfrac{1}{4 \pi \epsilon _0} \dfrac{(dq)x}{{(x^2 + y^2)}^{3/2}} \\\\ \implies \sf dE = \dfrac{1}{4 \pi \epsilon _0} \dfrac{\sigma 2 \pi ydy x}{{(x^2 + y^2)}^{3/2}}$

(Note: Distance of point P from the centre of disc is constant i.e. x)

$E = E_{axis} = \displaystyle \int^R_0 dE$

$\implies \sf E = \displaystyle \int dE = \dfrac{1}{4 \pi \epsilon _0} \displaystyle \int^R_0 dE \dfrac{\sigma 2 \pi xydy }{{(x^2 + y^2)}^{3/2}} \\ \\ \implies \sf E = \dfrac{2\sigma \pi x}{4 \pi \epsilon _0} \displaystyle \int^R_0 \dfrac{ ydy }{{(x^2 + y^2)}^{3/2}} \\\\ \implies \sf E = \dfrac{\sigma x}{2 \epsilon _0} \Bigg[-\dfrac{1}{\sqrt{x^2 + y^2}} \bigg|_{y = 0}^{y = R} \Bigg] \\\\ \implies \sf E = \dfrac{\sigma }{2 \epsilon _0} \Bigg[1 - \dfrac{x}{\sqrt{x^2 + R^2}} \Bigg]$

Answer 2

Answer:

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Explanation:

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