Explain electrostatic potential energy difference and give the noteworthy comments on it.
Explain electrostatic potential energy difference and give the noteworthy comments on it.
At every point in electric field a particle with charge $q$ possesses a certain electrostatic potential energy this work done increases its potential energy by an amount equal to the potential energy difference between points $R$ and $P$.
Thus, potential energy difference,
$U_{\mathrm{P}}-U_{\mathrm{R}}$
$\therefore \Delta U=U_{\mathrm{P}}-U_{\mathrm{R}}$
$\therefore \Delta U=W_{\mathrm{RP}}$
Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge $q$ from one point to another for electric field of any arbitrary charge configuration.
Following comments may be made :
$(i)$ The right side of equation $(1)$ depends only on the initial and final positions of the charge.
- It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force.
$(ii)$ The actual value of potential energy is not significant it is only the difference of potential energy that is significant.
The potential energy difference,
$\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{R}}$ $\therefore \Delta \mathrm{U} =\mathrm{U}_{\mathrm{P}}-\mathrm{U}_{\mathrm{R}}$ $\therefore \Delta \mathrm{U} =\mathrm{W}_{\mathrm{RP}}$
If potential energy is zero at infinity and adding an arbitrary constant $\alpha$ to potential energy at every point then,
$\left(\mathrm{U}_{\mathrm{P}}+\alpha\right)-\left(\mathrm{U}_{\mathrm{R}}+\alpha\right)=\mathrm{U}_{\mathrm{P}}-\mathrm{U}_{\mathrm{R}}$
$\therefore\left(\mathrm{U}_{\mathrm{P}}+\alpha-\mathrm{O}-\alpha\right)=\mathrm{U}_{\mathrm{P}}-\mathrm{U}_{\mathrm{R}}$
$\therefore\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{P}}-\mathrm{U}_{\mathrm{R}}$
The work done for bringing a charge from point $\mathrm{R}$ to $\mathrm{P}$ at infinity distance,
$\mathrm{W}_{\mathrm{RP}}=\mathrm{U}_{\mathrm{P}} \text { or } \mathrm{W}_{\infty \mathrm{P}}=\mathrm{U}_{\mathrm{P}}$
Since above equation provides a definition of potential energy of charge at any point $P$. "Potential energy of charge $q$ at a point is the work done by the external force (equal and opposite to the electric field) in bringing the charge $q$ from infinity to that point".
Similar Questions
The diagram shows a small bead of mass $m$ carrying charge $q$. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge $+Q$ has also been fixed as shown. The potential atthe point $P$ due to $+Q$ is $V$. The velocity with which the bead should projected from the point $P$ so that it can complete a circle should be greater than
The diagram shows a small bead of mass $m$ carrying charge $q$. The bead can freely move on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge $+Q$ has also been fixed as shown. The potential atthe point $P$ due to $+Q$ is $V$. The velocity with which the bead should projected from the point $P$ so that it can complete a circle should be greater than
Three identical small electric dipoles are arranged parallel to each other at equal separation a as shown in the figure. Their total interaction energy is $U$. Now one of the end dipole is gradually reversed, how much work is done by the electric forces.
Three identical small electric dipoles are arranged parallel to each other at equal separation a as shown in the figure. Their total interaction energy is $U$. Now one of the end dipole is gradually reversed, how much work is done by the electric forces.
A charged particle of charge $Q $ is held fixed and another charged particle of mass $m$ and charge $q$ (of the same sign) is released from a distance $r.$ The impulse of the force exerted by the external agent on the fixed charge by the time distance between $Q$ and $q$ becomes $2r$ is
A charged particle of charge $Q $ is held fixed and another charged particle of mass $m$ and charge $q$ (of the same sign) is released from a distance $r.$ The impulse of the force exerted by the external agent on the fixed charge by the time distance between $Q$ and $q$ becomes $2r$ is
A particle of mass $m$ having negative charge $q$ move along an ellipse around a fixed positive charge $Q$ so that its maximum and minimum distances from fixed charge are equal to $r_1$ and $r_2$ respectively. The angular momentum $L$ of this particle is
A particle of mass $m$ having negative charge $q$ move along an ellipse around a fixed positive charge $Q$ so that its maximum and minimum distances from fixed charge are equal to $r_1$ and $r_2$ respectively. The angular momentum $L$ of this particle is
A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ in the $x-$direction, where $a$ and $E _{0}$ are constants. Initially the particle was at rest at $x=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ in the $x-$direction, where $a$ and $E _{0}$ are constants. Initially the particle was at rest at $x=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
- [JEE MAIN 2020]