$A$ semicircular ring of radius $0.5 \ m$ is uniformly charged with a total charge of $1.4 \times 10^{-9} \ C$. The electric field intensity at the center of the ring is ........ $V/m$.

  • A
    $0$
  • B
    $320$
  • C
    $64$
  • D
    $32$

Explore More

Similar Questions

$A$ thin conducting ring of radius $R$ is given a charge $+Q.$ The electric field at the centre $O$ of the ring due to the charge on the part $AKB$ of the ring is $E.$ The electric field at the centre due to the charge on the part $ACDB$ of the ring is

Two charges, each equal to $-q$, are kept at $(-a, 0)$ and $(a, 0)$. $A$ charge $q$ is placed at the origin. If $q$ is given a small displacement $y$ along the $y$-direction, the force acting on $q$ is proportional to:

Four charges $+q, +2q, +q$ and $-2q$ are placed at the corners $A, B, C$ and $D$ of a square respectively. The force on a unit positive charge kept at the centre $O$ is

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

The diagram shows symmetrically placed rectangular insulators with uniformly charged distributions of equal magnitude. At the origin,the net electric field $\vec{E}_{net}$ is:

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