A thin disc of radius $b=2a$ has a concentric hole of radius $a$ in it (see figure). It carries uniform surface charge $\sigma$ on it. If the electric field on its axis at a height $\text{h}(\text{h}<<\text{a})$ from its centre is given as $\text{C}\text{h}$ then the value of $C$ is ![JEE Main 2015 Physics, Electrostatics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a64417cc3fb48d65f3f/question_1__q_648b5a64417cc3fb48d65f3f__cdn-question-pool.getmarks.app__image24-293f48f3-e45c-4786-87cd-28362409a1c3__215b242f13_final_ppt_sync.jpg)

$\because$ At the axial point of a uniformly charged disc electric field is given by, $E=\frac{\sigma }{2{\epsilon }_{0}}(1-cos\theta )$ ![JEE Main 2015 Physics, Electrostatics — solution figure 1](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a64417cc3fb48d65f3f/solution_1__q_648b5a64417cc3fb48d65f3f__cdn-question-pool.getmarks.app__image25-118f4e4e-ad3f-43b1-84f3-8d236828c2cc__f7ba2d6de9_final_ppt_sync.jpg) By superposition principle, when inner disc is removed , then, electric field due to remaining disc is, ![JEE Main 2015 Physics, Electrostatics — solution figure 2](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Electrostatics/648b5a64417cc3fb48d65f3f/solution_2__q_648b5a64417cc3fb48d65f3f__cdn-question-pool.getmarks.app__image26-66eba4b6-ba0e-4225-bad4-71e3fc8f7cf1__de14356be8_final_ppt_sync.jpg) $E=\frac{\sigma }{2{\epsilon }_{0}} [(1-cos{\theta }_{2})-(1-cos{\theta }_{1})]$ $=\frac{\sigma }{2{\epsilon }_{0}}[cos{\theta }_{1}-cos{\theta }_{2}]$ $=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{\sqrt{{h}^{2}+{a}^{2} }}-\frac{h}{\sqrt{{h}^{2}+{b}^{2}}}]$ $=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a\sqrt{1+\frac{{h}^{2}}{{a}^{2}}}}-\frac{h}{b\sqrt{1+\frac{{h}^{2}}{{b}^{2}} }}]$ $\because h \ll a$ and b. $\therefore E=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{b}]$ $=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{2a}]=\frac{\sigma h}{4{\epsilon }_{0}a}$ $\Rightarrow C=\frac{\sigma }{4{a\epsilon }_{0}}$.

A thin disc of radius b=2a has… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Electromagnetism

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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 $\sigma$ on it. If the electric field on its axis at a height $\text{h}(\text{h}<<\text{a})$ from its centre is given as $\text{C}\text{h}$ then the value of $C$ is

JEE Main 2015 Physics, Electrostatics — question figure

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
$\frac{\sigma }{4 a{\epsilon }_{0}}$
B
$\frac{\sigma }{a{\epsilon }_{0}}$
C
$\frac{\sigma }{5a{\epsilon }_{0}}$
D
$\frac{\sigma }{2a{\epsilon }_{0}}$

Did you get this right?

Solution

$\because$ At the axial point of a uniformly charged disc electric field is given by,

$E=\frac{\sigma }{2{\epsilon }_{0}}(1-cos\theta )$

JEE Main 2015 Physics, Electrostatics — solution figure 1

By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,

JEE Main 2015 Physics, Electrostatics — solution figure 2

$E=\frac{\sigma }{2{\epsilon }_{0}} [(1-cos{\theta }_{2})-(1-cos{\theta }_{1})]$

$=\frac{\sigma }{2{\epsilon }_{0}}[cos{\theta }_{1}-cos{\theta }_{2}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{\sqrt{{h}^{2}+{a}^{2} }}-\frac{h}{\sqrt{{h}^{2}+{b}^{2}}}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a\sqrt{1+\frac{{h}^{2}}{{a}^{2}}}}-\frac{h}{b\sqrt{1+\frac{{h}^{2}}{{b}^{2}} }}]$

$\because h \ll a$ and b.

$\therefore E=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{b}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{2a}]=\frac{\sigma h}{4{\epsilon }_{0}a}$

$\Rightarrow C=\frac{\sigma }{4{a\epsilon }_{0}}$ .

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Solution

$\because$ At the axial point of a uniformly charged disc electric field is given by,

$E=\frac{\sigma }{2{\epsilon }_{0}}(1-cos\theta )$

JEE Main 2015 Physics, Electrostatics — solution figure 1

By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,

JEE Main 2015 Physics, Electrostatics — solution figure 2

$E=\frac{\sigma }{2{\epsilon }_{0}} [(1-cos{\theta }_{2})-(1-cos{\theta }_{1})]$

$=\frac{\sigma }{2{\epsilon }_{0}}[cos{\theta }_{1}-cos{\theta }_{2}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{\sqrt{{h}^{2}+{a}^{2} }}-\frac{h}{\sqrt{{h}^{2}+{b}^{2}}}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a\sqrt{1+\frac{{h}^{2}}{{a}^{2}}}}-\frac{h}{b\sqrt{1+\frac{{h}^{2}}{{b}^{2}} }}]$

$\because h \ll a$ and b.

$\therefore E=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{b}]$

$=\frac{\sigma }{2{\epsilon }_{0}}[\frac{h}{a}-\frac{h}{2a}]=\frac{\sigma h}{4{\epsilon }_{0}a}$

$\Rightarrow C=\frac{\sigma }{4{a\epsilon }_{0}}$ .

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Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism