Three charges $Q, +q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic potential energy of the system is zero,the value of $Q$ is

  • A
    $\frac{-2q}{2+\sqrt{2}}$
  • B
    $\frac{+q}{2+\sqrt{2}}$
  • C
    $\frac{+2q}{2+\sqrt{2}}$
  • D
    $\frac{-q}{2+\sqrt{2}}$

Explore More

Similar Questions

Nine point charges are placed on a cube as shown in the figure. The charge $q$ is placed at the body centre whereas all other charges are at the vertices. The electrostatic potential energy of the system will be

Two charges $5 \text{ nC}$ and $-2 \text{ nC}$ are placed at points $(5 \text{ cm}, 0, 0)$ and $(23 \text{ cm}, 0, 0)$ in a region of space where there is no other external field. The electrostatic potential energy of this charge system is

If $OP = 1 \, cm$ and $OS = 2 \, cm$,calculate the work done by the electric field in shifting a point charge $q = \frac{4\sqrt{2}}{27} \, \mu C$ from point $P$ to $S$ in the given figure. The dipole moment is $p = 2 \times 10^{-6} \, C \cdot m$.

What is the potential energy of two equal positive point charges of $1\,\mu C$ each,held $1\, m$ apart in air?

Two point charges $q_1 = 6 \mu C$ and $q_2 = 4 \mu C$ are kept at points $A$ and $B$ in air where $AB = 10 \ cm$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \ cm$ (in $J$)?
$\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ SI units}\right)$

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