question_answer

A thin disc of radius b = 2a has a concentric hole of radius \['\sigma '\] in it (see figure). It carries uniform surface charge 'w' on it. If the electric field on its axis at height 'h' (h<<a) from its centre is given as \['C\overset{\scriptscriptstyle\smile}{h}'\]then value  of ‘C’ is:

[JEE ONLINE 10-04-2015]

A) \[\frac{\sigma }{a{{\varepsilon }_{0}}}\]

B) \[\frac{\sigma }{4a{{\varepsilon }_{0}}}\]

C) \[\frac{\sigma }{2a{{\varepsilon }_{0}}}\]

D) \[\frac{\sigma }{8a{{\varepsilon }_{0}}}\]

Correct Answer: B

Solution :

[b] \[\int_{{}}^{{}}{\overrightarrow{dE}}=\int_{{}}^{{}}{\frac{Kh\sigma 2\pi xdx}{{{\left( {{h}^{2}}+{{x}^{2}} \right)}^{3/2}}}}\]

\[E=\sigma Kh2\pi \int_{{}}^{{}}{\frac{xdx}{{{\left( {{h}^{2}}+{{x}^{2}} \right)}^{3/2}}}}\]      ..(i)

\[I=\int_{{}}^{{}}{\frac{xdx}{{{\left( {{h}^{2}}+{{x}^{2}} \right)}^{3/2}}}}\]        \[x=h\tan \theta \]

\[dx=h{{\sec }^{2}}\theta d\theta \]

\[I=\int_{{}}^{{}}{\frac{h\tan \theta \times h{{\sec }^{2}}\theta dx}{{{h}^{3}}{{\sec }^{3}}\theta }}\]

\[=\int_{{}}^{{}}{\frac{1}{h}\sin \theta d\theta }\]                    \[=-\frac{\cos \theta }{h}\]

\[I=-\frac{1}{h}\left( \frac{h}{\sqrt{{{x}^{2}}+{{h}^{2}}}} \right)\]

Put it in equation (i)

\[E=\frac{\sigma h}{2{{\varepsilon }_{0}}}\left[ -\frac{1}{\sqrt{{{x}^{2}}+{{h}^{2}}}} \right]_{x=a}^{x=2a}\]

\[=\frac{\sigma h}{2{{\varepsilon }_{0}}}\left[ \frac{1}{a}-\frac{1}{2a} \right]\]

\[E=\frac{\sigma h}{4{{\varepsilon }_{0}}a}=ch\] \[\Rightarrow \] \[C=\frac{\sigma }{4{{\varepsilon }_{0}}A}\]