A hemispherical bowl of radius R is uniformly | vedclass

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

Question

(A) Consider a thin ring element on the hemispherical shell at an angle $	heta$ with the base,having width $R d	heta$ and radius $r = R \cos 	heta$

Vedclass · Accepted Answer

$\frac{\sigma R}{2\varepsilon_0}$

Vedclass · Answer

(A) Consider a thin ring element on the hemispherical shell at an angle $	heta$ with the base,having width $R d	heta$ and radius $r = R \cos 	heta$. The charge on this ring is $dq = \sigma (2\pi r) (R d	heta) = 2\pi \sigma R^2 \cos 	heta d	heta$.The distance of every point on this ring from point $A$ (at distance $x = R/2$ from center) can be found using the law of cosines. However,a simpler approach is to use the potential due to a charged spherical shell. The potential at any point inside a uniformly charged spherical shell is constant and equal to $V = \frac{\sigma R}{\varepsilon_0}$.For a hemispherical shell,the potential at the center $O$ is $V_O = \frac{\sigma R}{2\varepsilon_0}$.For a point $A$ at a distance $x = R/2$ from the center on the base,the potential is given by $V_A = \frac{\sigma R}{2\varepsilon_0}$. This is because the potential at any point on the base of a hemispherical shell is constant and equal to the potential at the center.

$A$ hemispherical bowl of radius $R$ is uniformly charged with surface charge density $\sigma$. Find the electric potential at point $A$,which lies on the base at a distance $R/2$ from the center $O$ of the base.

Similar Questions

$A$ hollow metallic sphere of radius $10 \; cm$ is charged such that the potential at its surface is $80 \; V$. The potential at the centre of the sphere would be (in $; V$)

$A$ particle has a mass $400$ times that of an electron and a charge double that of an electron. It is accelerated by a potential difference of $5 \ V$. If the particle was initially at rest,what will be its final kinetic energy in $eV$?

$A$ regular hexagon of side $9 \ cm$ has a charge $5 \mu C$ at each of its vertices. The electric potential at its centre is $..........V$. $(k = 9 \times 10^9 \ Nm^2 C^{-2})$

$A$ proton is accelerated through a potential difference of $1 \,V$. The kinetic energy $(KE)$ of the proton will be $....... \,eV$.

$A$ particle of mass $m$ and charge $q$ is accelerated through a potential difference of $V$ volt. Its energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module