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

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

Question

(A) The electrostatic potential energy $U$ of a system of three charges $q_1, q_2, q_3$ separated by distances $r_{12}, r_{23}, r_{31}$ is given by $U

Vedclass · Accepted Answer

$-\frac{q}{2}$

Vedclass · Answer

(A) The electrostatic potential energy $U$ of a system of three charges $q_1, q_2, q_3$ separated by distances $r_{12}, r_{23}, r_{31}$ is given by $U = \frac{1}{4\pi\epsilon_0} [\frac{q_1q_2}{r_{12}} + \frac{q_2q_3}{r_{23}} + \frac{q_3q_1}{r_{31}}]$.Here,the charges are $q_1 = +q$,$q_2 = +q$,and $q_3 = Q$. The distance between each pair is $a$.Substituting these values into the formula,we get:$U = \frac{K}{a} [q \cdot q + q \cdot Q + Q \cdot q] = 0$Since $K/a 
eq 0$,the term inside the bracket must be zero:$q^2 + 2qQ = 0$$2qQ = -q^2$$Q = -\frac{q^2}{2q} = -\frac{q}{2}$

The charges $+q$,$+q$,and $Q$ are placed at the vertices of an equilateral triangle of side $a$. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to:

Similar Questions

An electric dipole of moment $p$ is placed in the position of stable equilibrium in a uniform electric field of intensity $E$. It is rotated through an angle $\theta$ from the initial position. The potential energy of the electric dipole in the final position is

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:

When two electrons $(e^-)$ move towards each other,the electrostatic potential energy of the system will:

Identify the $WRONG$ statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module