A uniformly charged disc of radius R having s | vedclass

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Question

(A) To find the electric field at a distance $Z$ on the axis of a charged disc,consider an elemental ring of radius $r$ and thickness $dr$ on the disc

Vedclass · Accepted Answer

$E = \frac{\sigma}{2 \varepsilon_{0}} \left( 1 - \frac{Z}{(Z^{2} + R^{2})^{1/2}} \right)$

Vedclass · Answer

(A) To find the electric field at a distance $Z$ on the axis of a charged disc,consider an elemental ring of radius $r$ and thickness $dr$ on the disc.The area of this elemental ring is $dA = 2\pi r dr$.The charge on this elemental ring is $dq = \sigma dA = \sigma (2\pi r dr)$.The electric field $dE$ due to this ring at a point $Z$ on the axis is given by the formula for a charged ring:$dE = \frac{1}{4\pi \varepsilon_{0}} \frac{(dq) Z}{(Z^{2} + r^{2})^{3/2}} = \frac{1}{4\pi \varepsilon_{0}} \frac{(\sigma 2\pi r dr) Z}{(Z^{2} + r^{2})^{3/2}} = \frac{\sigma Z}{2 \varepsilon_{0}} \frac{r dr}{(Z^{2} + r^{2})^{3/2}}$.To find the total electric field $E$,integrate $dE$ from $r = 0$ to $r = R$:$E = \int_{0}^{R} \frac{\sigma Z}{2 \varepsilon_{0}} \frac{r dr}{(Z^{2} + r^{2})^{3/2}}$.Let $u = Z^{2} + r^{2}$,then $du = 2r dr$,so $r dr = \frac{du}{2}$.$E = \frac{\sigma Z}{4 \varepsilon_{0}} \int_{Z^{2}}^{Z^{2}+R^{2}} u^{-3/2} du = \frac{\sigma Z}{4 \varepsilon_{0}} \left[ \frac{u^{-1/2}}{-1/2} \right]_{Z^{2}}^{Z^{2}+R^{2}} = \frac{\sigma Z}{2 \varepsilon_{0}} \left[ -\frac{1}{\sqrt{u}} \right]_{Z^{2}}^{Z^{2}+R^{2}}$.$E = \frac{\sigma Z}{2 \varepsilon_{0}} \left( \frac{1}{Z} - \frac{1}{\sqrt{Z^{2} + R^{2}}} \right) = \frac{\sigma}{2 \varepsilon_{0}} \left( 1 - \frac{Z}{\sqrt{Z^{2} + R^{2}}} \right)$.

$A$ uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the $xy$-plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from the origin.

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