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}$

Answer 1

The Correct Option is 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}$