A thin disc of radius b = 2a has a ...
Question
A thin disc of radius b = 2a has a concentric hole of radius 'a' in it (see figure).It carries uniform surface charge 'a' on it. If the electric field on its axis at
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A thin disc of radius ( b=2 a ) has a concentric hole of radius 'a' in it (see figure). It carries uniform surface charge 'a' on it. If the electric field on its axis at height 'h' (h ( << ) a) from its centre is given as "Ch' then value of ' ( C ) ' is
(1) ( frac{sigma}{a varepsilon_{0}} )
(2) ( frac{sigma}{4 a varepsilon_{0}} )
(3) ( frac{sigma}{8 a varepsilon_{0}} )
(4) ( frac{sigma}{2 a varepsilon_{0}} )

JEE/Engineering Exams
Physics
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A thin disc of radius b = 2a has a concentric hole of radius 'a' in it (see figure).It carries uniform surface charge 'a' on it. If the electric field on its axis at height 'h' (h<< a) from its centre is given as "Ch' then value of 'C' is
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Electric field due to complete disc ( (mathrm{R}=2 mathrm{a}) )
( E_{1}=frac{sigma}{2 varepsilon_{0}}left[1-frac{x}{sqrt{R^{2}+x^{2}}}right] )
( E_{1}=frac{sigma}{2 varepsilon_{0}}left[1-frac{h}{sqrt{4 a^{2}+h^{2}}}right]=frac{sigma}{2 varepsilon_{0}}left[1-frac{h}{2 a}right] )
Electric field due to disc ( (mathrm{R}=mathrm{a}) )
( E_{2}=frac{sigma}{2 varepsilon_{0}}left(1-frac{h}{a}right) )
Electric field due to given disc. ( E=E_{1}-E_{2} )
( =frac{sigma h}{4 varepsilon_{0} a} )
Option (2) is correct answer.

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