Electrostatic Potential Energy

Electrostatic potential energy is a fundamental concept in physics, particularly when analyzing the behaviour of electric dipoles in an external electric field. A dipole, consisting of two equal and opposite charges separated by a distance, experiences a torque when placed in an electric field. This torque tends to align the dipole with the field, leading to a change in its potential energy. The concept of electrostatic potential energy is not just theoretical; it has real-life applications, such as in the design of capacitors, molecular chemistry, and even in understanding how water molecules align in electric fields, which plays a crucial role in various biological processes. In this article, we will Understand the behaviour of dipoles in electric fields helps us grasp the principles of electrostatic potential energy and related terms with solved examples.

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This Story also Contains

1. Electrostatic Potential Energy
2. Solved Examples Based On Electrostatics Potential Energy
3. Summary

Electrostatic Potential Energy

Electrostatic potential energy is the energy stored in a system of charged particles due to their positions relative to each other. It arises from the electrostatic forces between the charges and is a key concept in understanding how charged particles interact within an electric field. For a single charge in an electric field, the potential energy is given by U=qV, where q is the charge and V is the electric potential at the location of the charge. For multiple charges, the total electrostatic potential energy is the sum of the potential energies due to each pair of charges.

It is the amount of work done by external forces in bringing a body from $\mathrm{\infty }$ to a given point against electric force.

or It is defined as negative work done by the electric force in bringing a body from $\mathrm{\infty }$ to that point.

• It is a Scalar quantity

• SI Unit: Joule

• Dimension : $\left[M{L}^2{T}^{-2}\right]$

Electric Potential Energy at a Point

If the point charge Q is producing the electric field

The electric force on test charge q at a distance r from Q is given by $F=\frac{KQq}{r^2}$

The amount of work done by the electric force in bringing a test charge from $\mathrm{\infty }$ to r is given by

$W={\int }_{\mathrm{\infty }}^r\frac{KQq}{x^2}dx=-\frac{KQq}{r}$

And negative of this work is equal to electric potential energy

So $U=\frac{KQq}{r}$

$U\to$ electric potential energy
$r\to$ distance between two

Change of Potential Energy

if a charge q is moved from $r_1$ to $r_2$ in an electric field produced by charge Q

Then Change of potential energy is given as

$\begin{array}{rl}& \mathrm{\Delta }U=KQq[\frac{1}{r_2}-\frac{1}{r_1}]\\ & \mathrm{\Delta }U\to \text{ change of energy }\\ & r_1,r_2\to \text{ distances }\end{array}$

Potential Energy of System of Two Charge

$U=\frac{K{Q}_1{Q}_2}{r}(S.I{)}_{\text{where }}K=\frac{1}{4\pi {ϵ}_0}$

Potential Energy For a System of 3 Charges

$U=K(\frac{Q_1{Q}_2}{r_{12}}+\frac{Q_2{Q}_3}{r_{23}}+\frac{Q_1{Q}_3}{r_{13}})$

Work Energy Relation

$W=U_f-U_i$

Where W=work done by an external force

$\begin{array}{rl}& U_f-\text{ final }P.E\\ & U_i-\text{ initial P.E. }\end{array}$

The relation between Potential and Potential energy

$\begin{array}{rl}& U=\frac{KQq}{r}=q\left[\frac{KQ}{r}\right]\\ & \text{ As }\\ & \text{ But }=\frac{KQ}{r}\\ & \text{ So }U=qV\end{array}$

The potential is defined as Potential energy Per unit charge.

i.e $V=\frac{W}{Q}=\frac{U}{Q}$

Where $V\to$ Potential
$U\to$ Potential energy

Electron Volt

$1\mathrm{ev}=1.6\times {10}^{-19}\mathrm{J}=1.6\times {10}^{-12}\mathrm{erg}$.

It is the smallest practical unit of energy which is used in atomic and nuclear physics.

Electric potential Energy of Uniformly charged sphere

$U=\frac{3Q^2}{20\pi {ϵ}_{0}R}$

Where R is the radius and Q is - the total charge.

Energy density- It is defined as the energy stored for unit volume.

$U_v=\frac{U}{V}$

Where $U-$ Potential Energy and $V-$ Volume.

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Solved Examples Based On Electrostatics Potential Energy

Example 1: A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at the origin. A point charge q is moving towards the ring along the z-axis and has speed v at z=4a. The minimum value of v such that it crosses the origin is :

1) $\sqrt{\left(\frac{2}{m}\right)}{\left(\frac{4}{15}\frac{q^2}{4\pi {\epsilon }_{o}a}\right)}^{\frac{1}{2}}$
2) $\sqrt{\left(\frac{2}{m}\right)}{\left(\frac{1}{5}\frac{q^2}{4\pi {\epsilon }_{o}a}\right)}^{\frac{1}{2}}$
3) $\sqrt{\left(\frac{2}{m}\right)}{\left(\frac{2}{15}\frac{q^2}{4\pi {\epsilon }_{o}a}\right)}^{\frac{1}{2}}$
4) $\sqrt{\left(\frac{2}{m}\right)}{\left(\frac{1}{15}\frac{q^2}{4\pi {\epsilon }_{o}a}\right)}^{\frac{1}{2}}$

Solution:

E and V at a point P that lies on the axis of the ring

$E_x=\frac{kQx}{{(x^2+R^2)}^{\frac{3}{2}}}V=\frac{kQ}{{(x^2+R^2)}^{\frac{1}{2}}}$

Use energy conservation

$\begin{array}{rl}& \mathrm{\Delta }K\cdot E+\mathrm{\Delta }U=0\\ & 0-\frac{1}{2}m{v}^2=q(\frac{kq}{5a}-\frac{kq}{3a})\\ & \frac{1}{2}m{v}^2=\frac{2k{q}^2}{15a}\\ & v=\left(\frac{4}{15}\right)\left(\frac{k{q}^2}{am}\right)\\ & v=\sqrt{\frac{2}{m}}\times {(\frac{2}{15}\times \left(\frac{q^2}{4\pi {\epsilon }_o}\right)\times \frac{1}{a})}^{\frac{1}{2}}\end{array}$

Hence, the answer is the option (3).

Example 2: In moving from A to B along an electric field line, the electric field does $6.4\times {10}^{-19}\mathrm{J}$ of work on an electron. If ${\varphi }_1,{\varphi }_2$ are equipotential surfaces, then the potential difference $(V_C-V_A)$ is

1) -4V

2) 4V

3) Zero

4) 64V

Solution:

Potential energy Per unit charge

$V=\frac{W}{Q}=\frac{U}{Q}$
wherein
S.I unit is $\frac{J}{C}$.

Work done by the field
$\begin{array}{rl}& W=q(-dV)=-e(V_A-V_B)=-e(V_B-V_A)=e(V_C-V_A)\Rightarrow \left((V_B=V_C)\right)\\ \Rightarrow & (V_C-V_A)=\frac{W}{e}=\frac{6.4\times {10}^{-19}}{1.6\times {10}^{-19}}=4V\end{array}$

Hence, the answer is the option (2).

Example 3: In free space, a particle A of charge $1\mu \mathrm{C}$ is held fixed at a point P. Another particle B of the same charge and mass of $4\mu \mathrm{kg}$ is kept at a distance of 1mm from P. If B is released then its velocity at a distance of 9mm from p is: $[$ Take $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}]$

1) $1.0\times {10}^3\mathrm{m}/\mathrm{s}$
2) $3.0\times {10}^4\mathrm{m}/\mathrm{s}$
3) $2.0\times {10}^3\mathrm{m}/\mathrm{s}$
4) $1.5\times {10}^2\mathrm{m}/\mathrm{s}$

Solution:

Loss in potential energy = gain in kinetic energy

Apply energy conservation

$\begin{array}{rl}& K\times {10}^{-6}\times {10}^{-6}[\frac{1}{{10}^{-3}}-\frac{1}{9\times {10}^{-3}}]=\frac{1}{2}m{v}^2\\ & \Rightarrow 9\times {10}^9\times \frac{{10}^{-6}\times {10}^{-6}}{{10}^{-3}}\times \frac{8}{9}=\frac{1}{2}\times 4\times {10}^{-6}{V}^2\Rightarrow V=2\times {10}^3\mathrm{m}/\mathrm{s}\end{array}$

Hence, the answer is the option (3).

Example 4: If $4\times {10}^{20}\mathrm{eV}$ energy is required to move a charge of 0.25 coulomb between two points. Then what will be the potential difference between them?

1)178V

2) 256V

3)356V

4)None of these.

Solution:

By using,

$\mathrm{KE}=\mathrm{QV}\Rightarrow 4\times {10}^{-20}\times 1.6\times {10}^{-1}=0.25\times \mathrm{V}\Rightarrow \mathrm{V}=256\mathrm{volt}$

Hence, the answer is the option (2).

Summary

Electrostatic potential energy is a crucial concept in physics, representing the energy stored due to the relative positions of charged particles. It plays a significant role in understanding electric fields and potential energy changes, as illustrated by various examples. For instance, the potential energy changes due to the movement of charges in electric fields or the energy required to move charges between points are key applications of this concept. Practical examples, such as calculating the minimum velocity needed for a charge to cross a ring or determining potential differences based on work done, highlight its relevance in both theoretical and real-world scenarios.