The surface charge density of a thin charged | vedclass

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(A) The electric field at the centre of a disc is $E = \frac{\sigma}{2\epsilon_0}$.The electric field at a distance $x$ along the axis of a disc is gi

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reduces by $70.7\%$

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(A) The electric field at the centre of a disc is $E = \frac{\sigma}{2\epsilon_0}$.The electric field at a distance $x$ along the axis of a disc is given by $E' = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}ight)$.Given $x = R$,we substitute this into the formula:$E' = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{R^2 + R^2}}ight) = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{R}{\sqrt{2}R}ight) = E \left(1 - \frac{1}{\sqrt{2}}ight)$.Since $\frac{1}{\sqrt{2}} \approx 0.707$,we have $E' = E(1 - 0.707) = 0.293E$.The reduction in the electric field is $E - E' = E - 0.293E = 0.707E$.The percentage reduction is $\frac{0.707E}{E} 	imes 100\% = 70.7\%$.

The surface charge density of a thin charged disc of radius $R$ is $\sigma$. The value of the electric field at the centre of the disc is $\frac{\sigma}{2\epsilon_0}$. With respect to the field at the centre,the electric field along the axis at a distance $R$ from the centre of the disc:

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