Chapter 24: Electric Potential

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Electric potential is a cornerstone of electrostatics, providing a scalar approach to understanding electric fields, with applications like calculating energy in rocket propulsion systems. Building on the concepts of electric fields and Gauss’s law from Chapters 21–23, this chapter explores electric potential in depth. For JEE Main, JEE Advanced, and NEET students, mastering electric potential is essential, as it frequently appears in problems involving energy, field calculations, and capacitor systems. This chapter, Electric Potential, covers electric potential and potential energy, potential due to charge distributions, relation between electric field and potential, and potential in conductors and capacitors, providing detailed explanations, derivations, solved examples, and practical applications to ensure conceptual clarity and problem-solving proficiency.

24.1 Electric Potential and Potential Energy

Electric potential provides a scalar measure of the electric field, simplifying energy calculations, a key concept for JEE/NEET electrostatics problems.

Electric Potential Energy

The electric potential energy $U$ of a charge $q$ in an electric field is the work done by the field to bring the charge from infinity to its position:

$$U=qV$$

• $V$: Electric potential at the position (defined below).
• For two point charges $q_1$ and $q_2$ separated by $r$:

$$U=k\frac{q_1{q}_2}{r}$$

• Units: Joules (J).
• Sign: Positive if repulsive (like charges), negative if attractive (unlike charges).

Electric Potential

Electric potential $V$ at a point is the potential energy per unit charge:

$$V=\frac{U}{q}=k\frac{Q}{r}$$

• For a point charge $Q$, at distance $r$ from $Q$.
• $k=\frac{1}{4\pi {ϵ}_0}\approx 9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$, ${ϵ}_0=8.85\times {10}^{-12}{\text{C}}^2/{\text{N·m}}^2$.
• Units: Volts (V), where $1\text{V}=1\text{J/C}$.
• Reference: $V=0$ at infinity.
• For multiple charges: $V=\sum V_i=k\sum \frac{q_i}{r_i}$ (superposition principle).

Work Done by the Field

Work done by the field when a charge $q$ moves between points with potentials $V_a$ and $V_b$:

$$W=q(V_a-V_b)$$

Properties

• Scalar Quantity: Potential is a scalar, simplifying calculations compared to vector fields.
• Conservative Field: Work done by the field is path-independent, only depends on the potential difference.
• Sign: $V$ is positive for positive charges, negative for negative charges.

Derivation: Potential Due to a Point Charge
The electric field due to a point charge $Q$ is $\overrightarrow{E}=k\frac{Q}{r^2}\hat{r}$. Potential is defined as $V=-{\int }_{\mathrm{\infty }}^r\overrightarrow{E}\cdot d\overrightarrow{l}$, where the path is radial from infinity to $r$, $d\overrightarrow{l}=dr\hat{r}$, $\overrightarrow{E}\cdot d\overrightarrow{l}=Edr$.

$$V=-{\int }_{\mathrm{\infty }}^r\left(k\frac{Q}{r^{\prime 2}}\right)d{r}^{\prime }=-kQ{[-\frac{1}{r^{\prime }}]}_{\mathrm{\infty }}^r=-kQ(-\frac{1}{r}-0)=k\frac{Q}{r}$$

Derivation: Potential Energy of Two Charges
For charges $q_1$ and $q_2$ at distance $r$, bring $q_2$ from infinity to $r$ in the field of $q_1$. The potential at $r$ due to $q_1$ is $V=k\frac{q_1}{r}$, so $U=q_{2}V=k\frac{q_1{q}_2}{r}$.

Derivation: Work Done by the Field
Work done by the field is the negative change in potential energy: $W=-\mathrm{\Delta }U=-(U_b-U_a)=-q(V_b-V_a)=q(V_a-V_b)$.

Derivation: Potential in Rocket System
In a rocket ion engine, a charge $Q=2\times {10}^{-6}\text{C}$ at the origin creates $V=k\frac{Q}{r}$ at $r=0.1\text{m}$, $V=1.8\times {10}^5\text{V}$, affecting ion energy (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves calculating the potential at $r=0.2\text{m}$ from a charge $Q=3\mu \text{C}$.

• Solution:
  $V=k\frac{Q}{r}=9\times {10}^9\times \frac{3\times {10}^{-6}}{0.2}=1.35\times {10}^5\text{V}$.
  • JEE Tip: Potential is a scalar; $V\propto 1/r$, unlike $E\propto 1/r^2$. Common error: Using $E$ formula instead of $V$.

Solved Example: A NEET problem involves two charges $q_1=+2\mu \text{C}$ at $(0,0)$ and $q_2=-2\mu \text{C}$ at $(0.1,0)$. Find $V$ at $(0.05,0)$.

• Solution:
  $V_1=k\frac{q_1}{r_1}=9\times {10}^9\times \frac{2\times {10}^{-6}}{0.05}=3.6\times {10}^5\text{V}$, $V_2=9\times {10}^9\times \frac{-2\times {10}^{-6}}{0.05}=-3.6\times {10}^5\text{V}$, $V=V_1+V_2=0\text{V}$.
  • NEET Tip: Potentials add as scalars; symmetry can lead to cancellation. Common error: Adding magnitudes without considering signs.

Solved Example: A JEE Advanced problem involves the potential energy of $q_1=4\mu \text{C}$ and $q_2=-3\mu \text{C}$ at $r=0.2\text{m}$.

• Solution:
  $U=k\frac{q_1{q}_2}{r}=9\times {10}^9\times \frac{(4\times {10}^{-6})\times (-3\times {10}^{-6})}{0.2}=-0.54\text{J}$.
  • JEE Tip: Negative $U$ indicates attraction; include charge signs. Common error: Forgetting the sign of $U$.

Solved Example: A JEE Main problem involves a charge $q=2\mu \text{C}$ moving from $V_a=100\text{V}$ to $V_b=50\text{V}$. Find the work done by the field.

• Solution:
  $W=q(V_a-V_b)=(2\times {10}^{-6})\times (100-50)=1\times {10}^{-4}\text{J}$.
  • JEE Tip: Work depends on potential difference; positive work means field does work. Common error: Using incorrect $q$ units.

Application: Potential applies to capacitors, energy storage, and rocketry (e.g., ion energy in propulsion, aligning with your interest, April 19, 2025).

24.2 Electric Potential Due to Charge Distributions

Electric potential due to continuous charge distributions requires integration, a common technique in JEE/NEET problems involving lines, rings, and spheres.

General Method

For a charge distribution, the potential at a point is:

$$V=k\int \frac{dq}{r}$$

• $dq$: Infinitesimal charge element.
• $r$: Distance from $dq$ to the field point.
• Unlike electric field, potential is a scalar, so no vector components are needed.

Potential Due to a Line Charge

For a line charge with linear charge density $\lambda$ (C/m), length $L$, along the x-axis from $-L/2$ to $L/2$, potential at $(0,d)$:

$$V=k\lambda {\int }_{-L/2}^{L/2}\frac{dx}{\sqrt{x^2+d^2}}=k\lambda \ln \left(\frac{\sqrt{(L/2{)}^2+d^2}+L/2}{\sqrt{(L/2{)}^2+d^2}-L/2}\right)$$

Potential Due to a Ring of Charge

For a ring of radius $R$, total charge $Q$, in the xy-plane, potential on the z-axis at $(0,0,z)$:

$$V=k\frac{Q}{\sqrt{R^2+z^2}}$$

Potential Due to a Uniformly Charged Disk

For a disk of radius $R$, surface charge density $\sigma$, potential on the axis at distance $z$:

$$V=\frac{\sigma }{2{ϵ}_0}(\sqrt{R^2+z^2}-|z|)$$

Potential Due to a Spherical Shell

For a thin spherical shell, radius $R$, charge $Q$:

• Outside ($r>R$): $V=k\frac{Q}{r}$.
• Inside ($r<R$): $V=k\frac{Q}{R}$ (constant, since $E=0$ inside).

Derivation: Potential Due to a Line Charge
Element $dx$ at $(x,0)$, $dq=\lambda dx$, distance to $(0,d)$ is $r=\sqrt{x^2+d^2}$. Potential: $dV=k\frac{dq}{r}$, integrate: $V=k\lambda {\int }_{-L/2}^{L/2}\frac{dx}{\sqrt{x^2+d^2}}$, yielding the result above.

Derivation: Potential Due to a Ring
Ring in xy-plane, $dq=\frac{Q}{2\pi R}Rd\theta$. At $(0,0,z)$, distance $r=\sqrt{R^2+z^2}$, constant for all $dq$. Integrate: $V=k{\int }_0^{2\pi }\frac{dq}{\sqrt{R^2+z^2}}=k\frac{Q}{\sqrt{R^2+z^2}}$.

Derivation: Potential Due to a Spherical Shell
Outside: Same as a point charge, $V=k\frac{Q}{r}$. Inside: $E=0$, so $V$ is constant, equal to the surface value: $V=k\frac{Q}{R}$.

Derivation: Potential in Rocket Ion Engine
A charged disk in an ion engine ($R=0.1\text{m}$, $\sigma ={10}^{-5}{\text{C/m}}^2$) at $z=0.05\text{m}$: $V\approx 8.47\times {10}^4\text{V}$, influencing ion trajectories (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a line charge, $\lambda =2\times {10}^{-6}\text{C/m}$, $L=0.4\text{m}$, along the x-axis. Find $V$ at $(0,0.2)$.

• Solution:
  $V=k\lambda \ln \left(\frac{\sqrt{(0.2{)}^2+(0.2{)}^2}+0.2}{\sqrt{(0.2{)}^2+(0.2{)}^2}-0.2}\right)=9\times {10}^9\times (2\times {10}^{-6})\ln \left(\frac{0.4828}{0.0828}\right)\approx 3.24\times {10}^4\text{V}$.
  • JEE Tip: Integrate scalar contributions; no symmetry cancellation needed. Common error: Using $E$ formula.

Solved Example: A NEET problem involves a ring, $R=0.1\text{m}$, $Q=3\mu \text{C}$, at $z=0.1\text{m}$. Find $V$.

• Solution:
  $V=k\frac{Q}{\sqrt{R^2+z^2}}=9\times {10}^9\times \frac{3\times {10}^{-6}}{\sqrt{(0.1{)}^2+(0.1{)}^2}}=1.91\times {10}^5\text{V}$.
  • NEET Tip: Distance is constant for a ring; use the formula directly. Common error: Using incorrect $r$.

Solved Example: A JEE Advanced problem involves a disk, $R=0.2\text{m}$, $\sigma =5\times {10}^{-6}{\text{C/m}}^2$, at $z=0.1\text{m}$. Find $V$.

• Solution:
  $V=\frac{5\times {10}^{-6}}{2\times 8.85\times {10}^{-12}}(\sqrt{(0.2{)}^2+(0.1{)}^2}-0.1)\approx 1.05\times {10}^5\text{V}$.
  • JEE Tip: Disk potential requires integration; use the derived formula. Common error: Misapplying limits.

Solved Example: A JEE Main problem involves a spherical shell, $R=0.1\text{m}$, $Q=4\mu \text{C}$, at $r=0.05\text{m}$. Find $V$.

• Solution:
  Inside: $V=k\frac{Q}{R}=9\times {10}^9\times \frac{4\times {10}^{-6}}{0.1}=3.6\times {10}^5\text{V}$.
  • JEE Tip: Inside a shell, $V$ is constant; outside, $V\propto 1/r$. Common error: Assuming $V\propto 1/r$ inside.

Application: Charge distributions apply to capacitors, potential mapping, and rocketry (e.g., ion engine potential fields, aligning with your interest, April 19, 2025).

24.3 Relation Between Electric Field and Potential

The electric field and potential are intimately related, providing a powerful tool for JEE/NEET problems involving field calculations.

Relation

The electric field is the negative gradient of the potential:

$$\overrightarrow{E}=-\mathrm{\nabla }V=-(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k})$$

• In one dimension: $E_x=-\frac{dV}{dx}$.
• For a point charge, $V=k\frac{Q}{r}$, so $E=-\frac{d}{dr}\left(k\frac{Q}{r}\right)=k\frac{Q}{r^2}$, matching the field.

Potential Difference

The potential difference between two points is related to the field:

$$V_b-V_a=-{\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}$$

Equipotential Surfaces

Surfaces where $V$ is constant are equipotential. Since $\overrightarrow{E}=-\mathrm{\nabla }V$, $\overrightarrow{E}$ is perpendicular to equipotential surfaces.

Potential Energy and Field

For a system of charges, the field can be derived from the potential energy gradient: $\overrightarrow{F}=-\mathrm{\nabla }U$.

Derivation: Electric Field from Potential
Given $V=k\frac{Q}{r}$, in spherical coordinates, $V$ depends only on $r$, so $\mathrm{\nabla }V=\frac{\partial V}{\partial r}\hat{r}=-\frac{kQ}{r^2}\hat{r}$. Thus, $\overrightarrow{E}=-\mathrm{\nabla }V=k\frac{Q}{r^2}\hat{r}$, matching the field from Coulomb’s law.

Derivation: Potential Difference
Work done by the field is $W=q(V_a-V_b)$. Since $W={\int }_a^b\overrightarrow{F}\cdot d\overrightarrow{l}=q{\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}$, we have $V_a-V_b={\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}$, so $V_b-V_a=-{\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}$.

Derivation: Equipotential Surfaces
If $V$ is constant on a surface, $\mathrm{\nabla }V\perp$ to the surface (no change along the surface). Since $\overrightarrow{E}=-\mathrm{\nabla }V$, $\overrightarrow{E}$ is perpendicular to the equipotential surface.

Derivation: Field in Rocket System
A potential $V={10}^5\text{V}$ at a plate in an ion engine, with $V$ decreasing to $0$ over $0.1\text{m}$, gives $E\approx -\frac{\mathrm{\Delta }V}{\mathrm{\Delta }x}={10}^6\text{N/C}$, accelerating ions (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves $V=k\frac{2\times {10}^{-6}}{r}$. Find $E$ at $r=0.2\text{m}$.

• Solution:
  $E=-\frac{dV}{dr}=-\frac{d}{dr}\left(k\frac{2\times {10}^{-6}}{r}\right)=k\frac{2\times {10}^{-6}}{r^2}=9\times {10}^9\times \frac{2\times {10}^{-6}}{(0.2{)}^2}=4.5\times {10}^5\text{N/C}$.
  • JEE Tip: $E=-\frac{dV}{dr}$ for radial fields; confirm with Coulomb’s law. Common error: Forgetting the negative sign.

Solved Example: A NEET problem involves a uniform field $E=500\hat{i}\text{N/C}$. Find the potential difference between $(0,0)$ and $(0.1,0)$.

• Solution:
  $V_b-V_a=-{\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}=-500\times 0.1=-50\text{V}$.
  • NEET Tip: Potential decreases along the field direction; path is simple here. Common error: Incorrect sign of $\mathrm{\Delta }V$.

Solved Example: A JEE Advanced problem involves $V=1000-200x\text{V}$. Find $\overrightarrow{E}$.

• Solution:
  $\overrightarrow{E}=-\mathrm{\nabla }V=-\frac{\partial V}{\partial x}\hat{i}=-(-200)\hat{i}=200\hat{i}\text{N/C}$.
  • JEE Tip: Use partial derivatives for each direction; here, $V$ depends only on $x$. Common error: Missing negative sign in $\overrightarrow{E}$.

Solved Example: A JEE Main problem involves equipotential surfaces for a point charge. Describe their shape.

• Solution:
  $V=k\frac{Q}{r}$, so constant $V$ means constant $r$. Equipotentials are spherical surfaces centered on the charge. $\overrightarrow{E}$ is radial, perpendicular to these spheres.
  • JEE Tip: Equipotentials are perpendicular to $\overrightarrow{E}$; for a point charge, they’re spheres. Common error: Assuming equipotentials follow field lines.

Application: Field-potential relations apply to field mapping, capacitor design, and rocketry (e.g., ion acceleration in engines, aligning with your interest, April 19, 2025).

24.4 Electric Potential in Conductors and Capacitors

Electric potential in conductors and capacitors is a practical application, frequently tested in JEE/NEET problems involving energy storage and field behavior.

Potential in Conductors

• Equilibrium: Inside a conductor, $E=0$, so $V$ is constant (equipotential volume).
• Surface: All charge resides on the surface; $V$ just outside is $V=k\frac{Q}{R}$ for a spherical conductor.
• Cavity: If a cavity contains charge $q$, $V$ inside adjusts due to induced charges; otherwise, $V$ is constant.

Potential in Capacitors

For a parallel plate capacitor with charge $Q$, area $A$, separation $d$:

• Field: $E=\frac{\sigma }{{ϵ}_0}=\frac{Q}{{ϵ}_{0}A}$.
• Potential difference: $V=Ed=\frac{Qd}{{ϵ}_{0}A}$.
• Capacitance: $C=\frac{Q}{V}=\frac{{ϵ}_{0}A}{d}$.

Energy Stored in a Capacitor

Energy stored:

$$U=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV$$

Potential Energy in Systems

For a system of charges, total potential energy is the sum of pairwise interactions: $U=\sum _{i<j}k\frac{q_i{q}_j}{r_{ij}}$.

Derivation: Potential Inside a Conductor
Inside a conductor, $E=0$ (from Gauss’s law, Chapter 23). Since $\overrightarrow{E}=-\mathrm{\nabla }V$, if $E=0$, $\mathrm{\nabla }V=0$, so $V$ is constant throughout the conductor.

Derivation: Potential Difference in a Capacitor
For a parallel plate capacitor, $E=\frac{\sigma }{{ϵ}_0}$, constant between plates. $V=-{\int }_0^{d}Edz=Ed=\frac{\sigma d}{{ϵ}_0}=\frac{Qd}{{ϵ}_{0}A}$.

Derivation: Energy Stored in a Capacitor
Work to charge a capacitor: $dW=Vdq$, where $V=\frac{q}{C}$. Integrate: $U={\int }_0^Q\frac{q}{C}dq=\frac{1}{2}\frac{Q^2}{C}$. Since $Q=CV$, $U=\frac{1}{2}C{V}^2$.

Derivation: Capacitor in Rocket System
A capacitor in a rocket circuit ($A=0.01{\text{m}}^2$, $d=0.001\text{m}$) has $C=\frac{{ϵ}_{0}A}{d}\approx 8.85\times {10}^{-11}\text{F}$, $V={10}^4\text{V}$, storing $U=4.43\times {10}^{-3}\text{J}$, powering ion acceleration (your interest, April 19, 2025).

Solved Example: A JEE Main problem involves a spherical conductor, $R=0.1\text{m}$, $Q=5\mu \text{C}$. Find $V$ on the surface.

• Solution:
  $V=k\frac{Q}{R}=9\times {10}^9\times \frac{5\times {10}^{-6}}{0.1}=4.5\times {10}^5\text{V}$. Inside, $V$ is the same.
  • JEE Tip: $V$ is constant inside a conductor; surface potential as if charge at center. Common error: Assuming $V=0$ inside.

Solved Example: A NEET problem involves a parallel plate capacitor, $A=0.02{\text{m}}^2$, $d=0.002\text{m}$, $Q=2\times {10}^{-8}\text{C}$. Find $V$.

• Solution:
  $V=\frac{Qd}{{ϵ}_{0}A}=\frac{(2\times {10}^{-8})\times 0.002}{(8.85\times {10}^{-12})\times 0.02}\approx 226\text{V}$.
  • NEET Tip: Use $V=Ed$, where $E=\frac{\sigma }{{ϵ}_0}$. Common error: Incorrect ${ϵ}_0$ value.

Solved Example: A JEE Advanced problem involves a capacitor with $C=10\mu \text{F}$, $V=100\text{V}$. Find the energy stored.

• Solution:
  $U=\frac{1}{2}C{V}^2=\frac{1}{2}\times (10\times {10}^{-6})\times (100{)}^2=0.05\text{J}$.
  • JEE Tip: Use any form of the energy formula; ensure units match. Common error: Forgetting the factor of $\frac{1}{2}$.

Solved Example: A JEE Main problem involves three charges $q_1=q_2=q_3=2\mu \text{C}$ at vertices of an equilateral triangle, side $0.1\text{m}$. Find $U$.

• Solution:
  $U=3\times k\frac{q^2}{r}=3\times 9\times {10}^9\times \frac{(2\times {10}^{-6}{)}^2}{0.1}=1.08\text{J}$.
  • JEE Tip: Sum pairwise interactions; all pairs are identical here. Common error: Miscounting pairs.

Application: Conductors and capacitors apply to circuits, energy storage, and rocketry (e.g., capacitor-driven ion propulsion, aligning with your interest, April 19, 2025).

Summary and Quick Revision

• Potential Energy: $U=k\frac{q_1{q}_2}{r}$, units: J. Sign depends on charges.
• Electric Potential: $V=k\frac{Q}{r}$, units: V. For multiple charges: $V=\sum k\frac{q_i}{r_i}$. Work: $W=q(V_a-V_b)$.
• Charge Distributions: Line: $V=k\lambda \ln \left(\frac{\sqrt{(L/2{)}^2+d^2}+L/2}{\sqrt{(L/2{)}^2+d^2}-L/2}\right)$; Ring: $V=k\frac{Q}{\sqrt{R^2+z^2}}$; Shell: $V=k\frac{Q}{r}$ (outside), $V=k\frac{Q}{R}$ (inside).
• Field-Potential Relation: $\overrightarrow{E}=-\mathrm{\nabla }V$, $V_b-V_a=-{\int }_a^b\overrightarrow{E}\cdot d\overrightarrow{l}$. Equipotentials perpendicular to $\overrightarrow{E}$.
• Conductors and Capacitors: Inside conductor: $V$ constant. Capacitor: $V=\frac{Qd}{{ϵ}_{0}A}$, $C=\frac{{ϵ}_{0}A}{d}$, $U=\frac{1}{2}C{V}^2$.
• Applications: Energy storage, field mapping, ion propulsion.
• JEE/NEET Tips: Use scalar addition for $V$, compute $\overrightarrow{E}$ via $-\mathrm{\nabla }V$, check equipotential geometry, apply conductor properties, verify significant figures (April 14, 2025).
• SI Units: Potential (V), potential energy (J), capacitance (F), field (N/C), charge (C).

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