Question

At what distance along the central axis of a uniformly charged plastic disc of radius R is the magnitude of the electric field equal to one-half the magnitude of the field at the centre of the surface of the disc?

• A. $ \frac{ R}{ \sqrt 2} $
• B. $ \frac{ R}{ \sqrt 3} $
• C. $ \sqrt {2R}$
• D. $ \sqrt {3R}$
• E. None of these

Answer 1

Answer: (B)

At a point on the axis of uniformly charged disc at a distance x above the centre of the disc, the magnitude of the electric fieid is
E = $ \frac{ \sigma}{ 2 \varepsilon_0} \bigg [ 1 - \frac{ x}{ \sqrt{ x^2 + R^2 }} \bigg ] $
but $ E_c = \frac{ \sigma}{ 2 \varepsilon_0} \, such \, that, \, \frac{ E}{ E_c } = \frac{ 1}{2} $
Then, $ 1 - \frac{ x}{ \sqrt{ x^2 + R^2 }} = \frac{ 1}{ 2} $
or $ \frac{ x}{ \sqrt{ x^2 + R^2 }} = \frac{ 1}{ 2} $
Squaring both sides and multiplying by $ x^2 + R^2$ to obtain
$ x^2 = \frac{ x^2 }{ 4 } + \frac{ R^2 }{ 4 } $
Thus, $ x^2 = \frac{ R^2 }{ 3 } $
x = $ \frac{ R}{ \sqrt 3}$