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BITSAT 2016 Paper

Section: Physics

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Question : 3 of 150

Marks: +1, -0

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\varepsilon_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

reduces by 70.7%
reduces by 29.3%
reduces by 9.7%
reduces by 14.6%

Solution:

Electric field intensity at the centre of the disc

E=\frac{\sigma}{2\varepsilon_0}

(given)

Electric field along the axis at any distance x from thecentre of the disc

E'=\frac{\sigma}{2E_0}\left(1-\frac{x}{\sqrt{x^2+R^2}}\right)

From question, x = R (radius of disc)

\therefore E'=\frac{\sigma}{2E_0}\left(1-\frac{R}{\sqrt{R^2+R^2}}\right)

=\frac{\sigma}{2E_0}\left(\left(\frac{\sqrt{2}}{R}-R\right)\left(\sqrt{2}R\right)\right)

=\frac{4}{14}E

∴ % reduction in the value of electric field

=\frac{\left(E-\frac{4}{14}E\right)\times100}{E}=\frac{1000}{14}\%=70.7\%

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