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

Answer: (B)

$\frac{ R}{ \sqrt 3}$

Answer 2

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 = $\frac{ R}{ \sqrt 3}$