A) \[\frac{R}{\sqrt{2}}\]
B) \[\frac{R}{\sqrt{3}}\]
C) \[\sqrt{2}R\]
D) \[\sqrt{3}R\]
E) None of these
Correct Answer: B
Solution :
At a point on the axis of a uniformly charged disc at a distance\[x\]above the centre of the disc, 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 \[\frac{x}{\sqrt{{{x}^{2}}+R}}=\frac{1}{2}\] Squaring both sides and multiplying by\[{{x}^{2}}+{{R}^{2}}\] to obtain \[{{x}^{2}}=\frac{{{x}^{2}}}{4}+\frac{{{R}^{2}}}{4}\] Thus, \[{{x}^{2}}=\frac{{{R}^{2}}}{3}\] \[x=\frac{R}{\sqrt{3}}\]
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