$A$ deuteron and an $\alpha$-particle are separated by a distance of $1\,\mathring{A}$ in air. The magnitude of the electric field due to the deuteron at the position of the $\alpha$-particle is:
- A$0$
- B$2.88 \times 10^{11}\,\text{N/C}$
- C$1.44 \times 10^{11}\,\text{N/C}$
- D$5.76 \times 10^{11}\,\text{N/C}$
Explore More
Similar Questions
$A$ uniformly charged rod of length $4\,cm$ and linear charge density $\lambda = 30\,\mu C/m$ is placed as shown in the figure. Calculate the $x-$ component of the electric field at point $P$.
Medium
View SolutionThe electric field at the centre $O$ of a semicircle of radius $a$ having a linear charge density $\lambda$ is given by:
DifficultAIPMT 2000
View SolutionWhen a $5 C$ charge is kept in a uniform electric field,a force of $5000 N$ acts on it. Find the potential difference between two points in that field,separated by a distance of $1 cm$. (in $V$)
Two infinitely long thin straight wires having uniform linear charge densities $2 \lambda$ and $3 \lambda$ are arranged parallel to each other at a distance $R$ apart. The intensity at a point midway between them is
For a uniformly charged ring of radius $R$,the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then the value of $h$ is
DifficultJEE MAIN 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo