The surface charge density of a thin charged disc of radius R is σ… | JEE Main 2013 (25 Apr, Shift 1)

Q45JEE Main 2013 · 25 APR · SHIFT 1Medium

Electric Field & Field Lines

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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Answer: (a)

On the axis of a disc: $E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{z}{\sqrt{z^2+R^2}}\right)$ .
At $z = R$ : $E = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{1}{\sqrt2}\right) = 0.293\,\dfrac{\sigma}{2\varepsilon_0}$
The field falls to 29.3% of the centre value, a reduction of $100 - 29.3 = 70.7\%$ .
So the answer is (a) reduces by 70.7%.

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