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{{\varepsilon }_{0}}}\left[ 1-\frac{x}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right] $ 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 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}} $