$A$ ring of charge with radius $R = 0.5\, m$ having a gap of length $l = 0.02\, m$,carries a total charge of $Q = +1\, C$. The electric field at the centre is

  • A
    $2.31\times10^4\, N/C$
  • B
    $2.31\times10^8\, N/C$
  • C
    $1.6\times10^4\, N/C$
  • D
    $1.6\times10^8\, N/C$

Explore More

Similar Questions

$A$ ring of radius $R$ has charge $-Q$ distributed uniformly over it. The charge that should be placed at the center of the ring such that the electric field becomes zero at a point on the axis of the ring at a distance $R$ from the center of the ring will be:

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC^{-1}$. The distance $d$ between the point charges $A$ and $B$ is ............ $m$. (in $.0$)

What is the magnitude of a point charge due to which the electric field $30\,cm$ away has the magnitude $2\,N/C$ $[1/4\pi \varepsilon_0 = 9 \times 10^9\,N\cdot m^2/C^2]$?

$A$ uniformly charged semicircular arc of radius $r$ has a linear charge density $\lambda$. What is the electric field at its centre? ($\varepsilon_0$ is the permittivity of free space)

An infinitely long uniformly charged string is placed along the $z-$axis. Its linear charge density is $\lambda$. $A$ point charge $q$ is moved from position $(a, 0, 0)$ to $(2a, 0, 0)$. The work done by the external agent is:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo