Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta$ with the $x$-axis,then the angle $\theta$ is

  • A
    $\tan^{-1}(1/2)$
  • B
    $\tan^{-1}(1/4)$
  • C
    $\tan^{-1}(1)$
  • D
    $\tan^{-1}(0)$

Explore More

Similar Questions

The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now, if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$, the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .

$A$ circular wire loop of radius $1 \, cm$ carries a total charge $1 \times 10^{-6} \, C$ distributed uniformly over its length. If $0.01 \%$ of its length (circumference) is cut off, then the electric field at the centre of the loop due to the remaining wire is
$(\text{Take} \, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \, \text{SI unit})$

There is an electric field $E$ in the $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ$ with the $X$-axis is $4.0\,J$,what is the value of $E$ in $N/C$?

$A$ 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

$A$ charge produces an electric field of $1\, N/C$ at a point distant $0.1\, m$ from it. The magnitude of the charge 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