Electric field intensity at the centre of the disc. E=2ϵ0σ (given) Electric field along the axis at any distance x from the centre of the disc E′=2ϵ0σ(1−x2+R2x) From question, x=R (radius of disc) ∴E′=2ϵ0σ(1−R2+R2R)=2ϵ0σ(2R2R−R) =144E ∴% reduction in the value of electric field =E(E−144E)×100=141000%=70.7%
The surface charge density of a thin charged disc of radius R is σ. The value of the electric field at the centre of the disc is σ 2 ∈_0. With…
JEE Main 2013 — Physics Electromagnetism
The surface charge density of a thin charged disc of radius R is σ. The value of the electric field at the centre of the disc is 2∈0σ. With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc:
Held on 25 Apr 2013 · Verified 6 Jul 2026.
Options
- A
reduces by 70.7%
- B
reduces by 29.3%
- C
reduces by 9.7%
- D
reduces by 14.6%
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Electromagnetism in other years
More JEE Main Electromagnetism questions
A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{j}$. Its associated electric field vector is ______.
A plane electromagnetic wave is moving in free space with velocity $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\hat{j}$ is the unit vector along $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:
A charged particle moves in a uniform magnetic field. If the velocity is perpendicular to the field, the path of the particle is:
A monochromatic source of light operating at $15$ kW emits $2.5\times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h=6.6\times 10^{-34}$ J·s and $c=3\times 10^8$ m/s).
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10 \mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$. $\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]$
Other Physics topics for JEE Main
Keep practicing Electromagnetism
Work through every JEE Main Electromagnetism PYQ, year by year.