A uniformly charged disc of radius R having s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

Two parallel large thin metal sheets have equal surface charge densities $\sigma = 26.4 \times 10^{-12} \ C/m^2$ of the same sign. The electric field between these sheets is:

Two parallel large thin metal sheets have equal surface charge densities $(\sigma = 26.4 \times 10^{-12} \, C/m^2)$ of opposite signs. The electric field between these sheets is

The net electric field at point $P$ due to the segments $dq_1$ and $dq_2$ of a uniformly charged spherical shell is ...... ($C$ is the center of the shell.)

Two infinitely long thin straight parallel wires are kept at a perpendicular distance of $2R$, having uniform linear charge densities $+\lambda$ and $-\lambda$ respectively. The magnitude of the electric field at the midpoint between the two wires will be . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module