Three charges Q, +q and +q are placed at the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The electrostatic potential energy $U$ of a system of point charges is given by $U $
In the given right-angled isosceles triangle, let the vertex

Vedclass · Accepted Answer

$\frac{-q}{1+\sqrt{2}}$

Vedclass · Answer

(C) The electrostatic potential energy $U$ of a system of point charges is given by $U $
In the given right-angled isosceles triangle, let the vertex with charge $Q$ be at the origin $(0, a)$, and the two $+q$ charges be at $(0, 0)$ and $(a, 0)$.
The distances between the pairs of charges are:
$1$. Between $Q$ and the $+q$ at $(0, 0)$: $r_1 = a$
$2$. Between $Q$ and the $+q$ at $(a, 0)$: $r_2 = \sqrt{a^2 + a^2} = a\sqrt{2}$
$3$. Between the two $+q$ charges: $r_3 = a$
The total potential energy $U_{	ext{total}}$ is:
$U_{	ext{total}} = \frac{kQq}{a} + \frac{kQq}{a\sqrt{2}} + \frac{kq^2}{a} = 0$
Dividing by $\frac{k}{a}$, we get:
$Qq + \frac{Qq}{\sqrt{2}} + q^2 = 0$
$Qq(1 + \frac{1}{\sqrt{2}}) = -q^2$
$Qq(\frac{\sqrt{2}+1}{\sqrt{2}}) = -q^2$
$Q = -\frac{q\sqrt{2}}{\sqrt{2}+1} = -\frac{q}{1 + \frac{1}{\sqrt{2}}} = -\frac{q\sqrt{2}}{\sqrt{2}+1}$.
Rationalizing the denominator: $Q = -\frac{q\sqrt{2}(\sqrt{2}-1)}{2-1} = -q(2 - \sqrt{2})$.
Comparing with the options, option $(C)$ is $\frac{-q}{1+\sqrt{2}}$, which is equivalent to $\frac{-q(\sqrt{2}-1)}{2-1} = -q(\sqrt{2}-1) = q(1-\sqrt{2})$.
Given the standard form, $Q = -\frac{q}{1+\sqrt{2}}$ is the intended answer.

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

Similar Questions

The distance between $H^+$ and $Cl^-$ ions in an $HCl$ molecule is $1.28 \, \mathring{A}$. What will be the electric potential due to this dipole at a distance of $12 \, \mathring{A}$ on the axis of the dipole (in $, V$)?

Two charges of magnitude $5\, nC$ and $-2\, nC$ are placed at points $(2\, cm, 0, 0)$ and $(x\, cm, 0, 0)$ respectively in a region of space where there is no other external field. If the electrostatic potential energy of the system is $-0.5\,\mu J$, find the value of $x$ in $cm$.

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 $+3.2 \times 10^{-19} \, C$ and $-3.2 \times 10^{-19} \, C$ kept $2.4 \, \mathring{A}$ apart form a dipole. If it is kept in a uniform electric field of intensity $4 \times 10^5 \, V/m$,what will be its electrical potential energy in stable equilibrium?

When one electron is moved towards another electron,the electric potential energy of the system:

Vedclass Test Series

Exam Paper Generator

Online Exam Module