Explain the positions (orientations) of a dipole in stable equilibrium,unstable equilibrium,and when the potential energy is zero.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The potential energy $U$ of a dipole placed in a uniform electric field $\vec{E}$ is given by $U = -\vec{p} \cdot \vec{E} = -pE \cos \theta$,where $\theta$ is the angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$.
$(i)$ Stable Equilibrium: When $\vec{p}$ is parallel to $\vec{E}$,$\theta = 0^{\circ}$.
$U = -pE \cos(0^{\circ}) = -pE$. This is the minimum potential energy state,representing stable equilibrium.
(ii) Unstable Equilibrium: When $\vec{p}$ is anti-parallel to $\vec{E}$,$\theta = 180^{\circ}$.
$U = -pE \cos(180^{\circ}) = pE$. This is the maximum potential energy state,representing unstable equilibrium.
(iii) Zero Potential Energy: When $\vec{p}$ is perpendicular to $\vec{E}$,$\theta = 90^{\circ}$.
$U = -pE \cos(90^{\circ}) = 0$. This represents the state where the potential energy of the dipole is zero.

Explore More

Similar Questions

As shown in the figure,a particle $A$ of mass $2m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q$. The system is placed in an electric field $\vec E$. The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E$. After the system settles into equilibrium,one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find the maximum tension in the rod.

For an electric dipole with charge $q = 2 \times 10^{-6} \ C$ and separation distance $d = 0.01 \ m$,if the external electric field is $E = 5 \times 10^{5} \ N/C$,calculate the maximum torque acting on the dipole.

An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity of $2 \times 10^5 \, \text{NC}^{-1}$. It experiences a torque equal to $4 \, \text{Nm}$. The charge on the dipole,if the dipole length is $2 \, \text{cm}$,is:

Charges $-q$ and $+q$ located at $A$ and $B$,respectively,constitute an electric dipole. Distance $AB = 2a$,$O$ is the midpoint of the dipole and $OP$ is perpendicular to $AB$. $A$ charge $Q$ is placed at $P$ where $OP = y$ and $y >> 2a$. The charge $Q$ experiences an electrostatic force $F$. If $Q$ is now moved along the equatorial line to $P'$ such that $OP' = \frac{y}{3}$,the force on $Q$ will be close to: $\left( \frac{y}{3} >> 2a \right)$

An electric dipole consists of two particles, each of mass $1 \ kg$, separated by $1 \ m$, carrying charges $1 \ \mu C$ and $-1 \ \mu C$ respectively. It is in equilibrium in a uniform electric field of $2 \times 10^4 \ Vm^{-1}$. If it is deflected by a small angle $2^{\circ}$, the minimum time taken by it to come back again to the mean position is (in seconds): (in $\pi$)

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