Explain electrostatic potential energy difference and give the noteworthy comments on it.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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".

898-s49

Similar Questions

Electrostatic potential energy of given system will be

There is an electric field $E$ in $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ $ with the $X$-axis is $4.0\;J$, what is the value of $E$........ $N/C$

  • [AIPMT 1995]

When a charge of $3\, coulomb$ is placed in a uniform electric field, it experiences a force of $3000\, Newton$. Within this field, potential difference between two points separated by a distance of $1\, cm$ is........$volts$

A test charge $q$ is made to move in the electric field of a point charge $Q$ along two different closed paths as per figure. First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases ?

$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \;\mu C$ and $-2\; \mu C$ (and with no external field) placed at $(-9 \;cm , 0,0)$ and $(9\; cm , 0,0)$ respectively.

$(b)$ How much work is required to separate the two charges infinitely away from each other?

$(c)$ Suppose that the same system of charges is now placed in an external electric field $E=A\left(1 / r^{2}\right) ; A=9 \times 10^{5} \;C m ^{-2} .$ What would the electrostatic energy of the configuration be?