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

In the following diagram,the work done in moving a point charge from point $P$ to points $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. Then ($A, B, C$ are points on a semicircle and point charge $q$ is at the centre of the semicircle):

Let $W$ joule be the work done to move an electric charge $q$ coulomb from a place $A$,where potential is $-5 \ V$,to another place $B$,where potential is $V$ volt. The value of $V$ is

$125$ identical charged small spheres coalesce to form a big charged sphere. If the electric potential on each small sphere is $60 \text{ mV}$, then the electric potential on the bigger sphere formed is: (in $\text{ V}$)

Two charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively, which are at a distance $2L$ apart. $C$ is the midpoint of $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CSD$ $(W_1)$ and along the line $CBD$ $(W_2)$ are:

The potential to which a conductor is raised,depends on

Vedclass Test Series

Exam Paper Generator

Online Exam Module