Question

At what distance along the central axis of a uniformly charged plastic disk 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 disk?

• A. $\frac{R}{\sqrt{2}}$
• B. $\frac{R}{\sqrt{3}}$
• C. $\sqrt{2}R$
• D. $\sqrt{3}R$

Answer 1

Answer: (B)

At a point on the axis of a uniformly charged disk at a distance x above the centre of the disk, the magnitude of the electric field is $E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{\sqrt{x^2+R^2}}\right]$ But $E_c=\frac{\sigma }{2{\epsilon }_0}$ 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 side and multiply by $x^2+R$ to obtain $x^2=\frac{x^2}{4}+\frac{R^2}{4}$ Thus, $x^2=\frac{R^2}{3}$ $x=\frac{R}{\sqrt{3}}$