Electric charge is transferred to an irregula | vedclass

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(C) For a charged conductor,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $(ROC)$ at that point,i.e.,$\sigm

Vedclass · Accepted Answer

$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$

Vedclass · Answer

(C) For a charged conductor,the surface charge density $\sigma$ is inversely proportional to the radius of curvature $(ROC)$ at that point,i.e.,$\sigma \propto \frac{1}{ROC}$.From the figure,the radius of curvature at the sharpest point (top) is the smallest,and it increases as we move towards the flatter sides.Comparing the points:$1$. The point at $\sigma_1$ has the smallest radius of curvature.$2$. The point at $\sigma_3$ has a larger radius of curvature than $\sigma_1$ but smaller than the sides.$3$. The points at $\sigma_2$ and $\sigma_4$ are on the flatter sides,where the radius of curvature is the largest and equal.Thus,the order of radius of curvature is $(ROC)_1 Since $\sigma \propto \frac{1}{ROC}$,the order of surface charge densities is $\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$.Therefore,the correct option is $(C)$.

Electric charge is transferred to an irregular metallic disk as shown in the figure. If $\sigma_1, \sigma_2, \sigma_3$ and $\sigma_4$ are surface charge densities at the given points,then choose the correct answer from the options given below:

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