A hemispherical bowl of radius R is uniformly | vedclass

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(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

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

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(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.

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