JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: Potential and Energy
TOPIC 5 OF 38
Potential and Energy
🎓 Class 12
Physics
CBSE
Theory
Ch 2 – Electrostatic Potential and Capacitance
⏱ ~14 min
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Potential and Energy
2.1 Introduction — Energy in the Electric Field
Lift a stone above the floor and release it — gravity does work, converting gravitational potential energy into kinetic energy. A strikingly parallel story unfolds around every electric charge. The Coulomb force is conservative, exactly like gravity, so we can define a scalar function — the electrostatic potential — whose difference between two points equals the work required per unit charge to move a test charge from one point to the other.
What you will learn in Part 1: The idea of electrostatic potential \(V\), its formula for point charges, dipoles and systems, the geometry of equipotential surfaces, and how to compute the potential energy stored in a collection of charges.
2.2 Electrostatic Potential
Imagine a stationary source charge producing an electric field \(\vec E\). If we bring a very small test charge \(q\) from infinity to a point P, the external agent must do work against the field. The work done per unit positive test charge is a property of the point P alone — it does not depend on \(q\). This ratio is defined as the electrostatic potential at P.
Definition. \[V_{\text{P}} = \frac{W_{\infty\to\text{P}}}{q}\] SI unit: volt (V) = 1 J/C. Dimensions: \([ML^2T^{-3}A^{-1}]\).
Only differences in potential have physical meaning; the zero of potential is a matter of convention (we pick infinity as the reference).
2.3 Potential Due to a Point Charge
Consider a point charge \(Q\) fixed at the origin. To find \(V\) at distance \(r\), compute the work done against \(\vec E\) in bringing a unit positive charge from \(\infty\) to the point:
\[V(r)=-\int_{\infty}^{r}\vec E\cdot d\vec r = -\int_{\infty}^{r}\frac{Q}{4\pi\varepsilon_0 r'^2}\,dr' = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}\]
Key formula: \(\displaystyle V = \frac{kQ}{r}\), with \(k = 9\times 10^9\) N·m\(^2\)/C\(^2\). For \(Q>0\), \(V>0\); for \(Q<0\), \(V<0\). Unlike the field, \(V\) is a scalar.
2.4 Potential Due to an Electric Dipole
A dipole consists of charges \(+q\) and \(-q\) separated by a small vector \(2\vec a\) pointing from \(-q\) to \(+q\). Its dipole moment is \(\vec p = q\cdot 2\vec a\). Let P be any point at distance \(r\) from the centre, making angle \(\theta\) with the dipole axis. For \(r \gg a\):
\[V(r,\theta)=\frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2}=\frac{1}{4\pi\varepsilon_0}\frac{\vec p\cdot\hat r}{r^2}\]
- Axial point (\(\theta=0\)): \(V = \dfrac{kp}{r^2}\) (maximum, positive).
- Other axial point (\(\theta=\pi\)): \(V = -\dfrac{kp}{r^2}\).
- Equatorial point (\(\theta=\pi/2\)): \(V = 0\) — the two charges are equidistant.
Unlike the potential of a monopole (which falls as \(1/r\)), a dipole's potential falls faster — as \(1/r^2\) — because distant observers see the two opposite charges nearly cancel.
2.5 Potential Due to a System of Charges
Because \(V\) is a scalar, the total potential at a point due to many charges is simply the algebraic sum:
\[V_{\text{P}} = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i}\]
where \(r_i\) is the distance from charge \(q_i\) to point P. No angle bookkeeping, no vector addition — a huge simplification compared with \(\vec E\).
2.6 Equipotential Surfaces
An equipotential surface is the set of all points at the same potential.
- Moving a charge along an equipotential needs zero work because \(\Delta V = 0\).
- The electric field is always perpendicular to the equipotential (any parallel component would do work).
- Two equipotential surfaces never intersect (a single point cannot have two potentials).
- Equipotentials are closer together where the field is stronger.
| Source | Equipotential Shape | Field-line Shape |
|---|---|---|
| Isolated point charge | Concentric spheres | Radial lines |
| Uniform field | Parallel planes \(\perp\) field | Parallel straight lines |
| Electric dipole | Curved, peanut-shaped surfaces | Curved from +q to -q |
| Two equal positive charges | Curved; flattens in between | Meet at a neutral point |
Relation Between Field and Potential
In moving a unit test charge through a small displacement \(d\vec l\) against the field, \(dW = -\vec E\cdot d\vec l = dV\). For displacement along the direction of decreasing potential we get:
\[E = -\frac{dV}{dr}\]
The field points from high to low potential; its magnitude equals the steepness (gradient) of the potential curve.
2.7 Potential Energy of a System of Charges
(a) Two charges in empty space
Bring \(q_1\) from infinity to its final position — no work is done because no other charge is present. Now bring \(q_2\) from infinity to a point at distance \(r_{12}\) from \(q_1\); the work required is \(q_2\times V_1(r_{12})\). The stored energy is therefore
\[U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}\]
Like charges: \(U>0\) (energy must be supplied to force them near). Unlike charges: \(U<0\) (energy released when they come together).
(b) Three or more charges
Add up the pair-wise energies — every pair is counted exactly once:
\[U = \frac{1}{4\pi\varepsilon_0}\left[\frac{q_1 q_2}{r_{12}}+\frac{q_1 q_3}{r_{13}}+\frac{q_2 q_3}{r_{23}}\right]\]
(c) Potential Energy in an External Field
If an external field \(\vec E_{\text{ext}}\) with potential \(V(\vec r)\) is already present, the energy of a single charge \(q\) placed at \(\vec r\) is simply
\[U = q\,V(\vec r)\]
For a system, the total energy = (interaction energy among the charges) + (energy of each charge with the external field):
\[U_{\text{total}} = \sum_{i
(d) Dipole in a Uniform External Field
When a dipole \(\vec p\) makes angle \(\theta\) with a uniform field \(\vec E\), its potential energy (taking \(U=0\) at \(\theta=\pi/2\)) is:
\[U(\theta)=-\vec p\cdot\vec E = -pE\cos\theta\]
Minimum at \(\theta=0\) (parallel, stable), maximum at \(\theta=\pi\) (anti-parallel, unstable).
Worked Examples — Potential and Energy
Example 2.1: Potential at a point
Find the potential at a point 30 cm from a point charge of \(+5\,\mu\)C in vacuum.
\[V = \frac{kQ}{r} = \frac{(9\times 10^9)(5\times 10^{-6})}{0.30} = \boxed{1.5\times 10^{5}\,\text{V}}\]
Example 2.2: Work done moving a charge between two potentials
How much work is required to carry a charge of \(2\,\mu\)C from a point at potential 200 V to a point at potential 600 V?
\[W = q\,\Delta V = (2\times 10^{-6})(600-200) = \boxed{8\times 10^{-4}\,\text{J}}\]
Since \(W>0\), the external agent does positive work (the charge is pushed uphill in potential).
Example 2.3: Potential at the centroid of an equilateral triangle
Three point charges \(+2\,\mu\)C, \(-3\,\mu\)C and \(+4\,\mu\)C are placed at the vertices of an equilateral triangle of side 20 cm. Find the potential at the centroid.
Distance from each vertex to centroid: \(r = \dfrac{a}{\sqrt 3} = \dfrac{0.20}{\sqrt 3} \approx 0.1155\) m.
\[V = \frac{k}{r}(q_1+q_2+q_3) = \frac{9\times 10^9}{0.1155}(2-3+4)\times 10^{-6}\]
\[= 7.79\times 10^{10}\times 3\times 10^{-6} = \boxed{2.34\times 10^{5}\,\text{V}}\]
Example 2.4: Electrostatic PE of three charges
Calculate the energy needed to assemble charges \(+1\,\mu\)C, \(+2\,\mu\)C, \(-3\,\mu\)C at the vertices of an equilateral triangle of side 10 cm.
\(r_{12}=r_{13}=r_{23}=0.10\) m.
\[U=\frac{k}{r}\left(q_1q_2+q_1q_3+q_2q_3\right)\]
\[=\frac{9\times 10^9}{0.10}\,[(1)(2)+(1)(-3)+(2)(-3)]\times 10^{-12}\]
\[=9\times 10^{10}\times(-7\times 10^{-12})=\boxed{-0.63\,\text{J}}\]
The negative sign shows the configuration is bound (energy is released on assembly).
Example 2.5: Dipole — potential on the axis
A dipole of moment \(p = 6\times 10^{-9}\) C·m points along \(+x\). Find the potential at a point 20 cm from its centre on the axis.
Axial: \(\theta=0\), \(\cos\theta=1\).
\[V=\frac{kp}{r^2}=\frac{(9\times 10^9)(6\times 10^{-9})}{(0.20)^2}=\boxed{1350\,\text{V}}\]
Example 2.6: Speed gained by a free electron
An electron is released from rest and accelerates through a potential difference of 100 V. Find its final kinetic energy and speed.
KE = \(eV=(1.6\times 10^{-19})(100)=1.6\times 10^{-17}\,\)J.
\[v=\sqrt{\tfrac{2\,\text{KE}}{m_e}}=\sqrt{\tfrac{2\times 1.6\times 10^{-17}}{9.1\times 10^{-31}}}=\boxed{5.93\times 10^{6}\,\text{m/s}}\]
Example 2.7: Potential midway between two charges
Charges \(+4\,\mu\)C and \(-2\,\mu\)C are placed 12 cm apart. Find the potential at the midpoint.
Midpoint is 6 cm from each charge.
\[V=\frac{k(q_1+q_2)}{r}=\frac{9\times 10^9 (4-2)\times 10^{-6}}{0.06}=\boxed{3\times 10^{5}\,\text{V}}\]
Activity — Mapping an EquipotentialL3 Apply
Predict: A pair of flat metal electrodes in a tray of weak salt water is connected to a 9 V battery. If you dip a voltmeter probe at many points in the tray, can you trace lines where the reading is the same? What shape will these lines take between the two electrodes?
- Pour a 3 mm layer of salt-water on a tray lined with graph paper.
- Place two rectangular copper strips as electrodes ~10 cm apart and connect to a 9 V battery.
- Keeping one voltmeter lead on the negative electrode, probe the tray and mark all points reading 3 V.
- Repeat for 6 V and 4.5 V. Join points of equal voltage with a smooth curve.
Observation: Equal-voltage points lie on smooth curves roughly parallel to the electrodes in the middle and curving around their ends. Spacing is smallest near the electrode edges (where the field is strongest).
Explanation: The salt-water carries a small steady current, but each instant the system looks like an electrostatic problem. Field lines run from the + electrode to the − electrode; equipotentials lie perpendicular to them. This is a direct experimental map of the 3-D electrostatic equipotential surfaces.
Explanation: The salt-water carries a small steady current, but each instant the system looks like an electrostatic problem. Field lines run from the + electrode to the − electrode; equipotentials lie perpendicular to them. This is a direct experimental map of the 3-D electrostatic equipotential surfaces.
Interactive: Potential Calculator L3 Apply
Enter up to three point charges (in μC) and their distances (in cm) from a field point P. The tool computes \(V = k\sum q_i/r_i\) at P.
Competency-Based Questions
During a science exhibition, a student sets up a small electrostatic demonstration using two point charges \(+6\,\mu\)C and \(-2\,\mu\)C placed 20 cm apart on a ruler. Probes are used to measure potential at various points around the ruler.
Q1. L1 Remember The SI unit of electrostatic potential is:
Answer: B. Volt = joule per coulomb.
Q2. L3 Apply Find the potential at the midpoint of the two demonstration charges. (3 marks)
Midpoint is 10 cm from each. \(V = \dfrac{9\times 10^9(6-2)\times 10^{-6}}{0.10} = \boxed{3.6\times 10^{5}\,\text{V}}\).
Q3. L2 Understand True/False: The potential at any point on the equatorial plane of the dipole formed by these two charges (if they were equal and opposite) would be zero.
True. On the equatorial plane, both charges are equidistant, so their scalar potentials add to zero.
Q4. L4 Analyse At what point on the line joining the two charges (outside the segment, on the side of the −2 μC charge) does the net potential equal zero? (3 marks)
Let the zero point be at distance \(x\) from the −2 μC charge, beyond it (i.e. distance \(x+0.20\) from +6 μC, but on the other side the zero lies between them first). Using \(\frac{6}{x+20}=\frac{2}{x}\) (cm): \(6x=2x+40 \Rightarrow x=10\) cm. So a zero-V point lies 10 cm from −2 μC, toward +6 μC (i.e. 10 cm inside the gap). A second zero occurs outside the segment on the +6 μC side at infinity only — so the only finite zero is at the midpoint-adjacent 10 cm location.
Q5. L3 Apply Compute the work required to move a \(+1\,\mu\)C test charge from the midpoint (Q2) to infinity. (2 marks)
\(W = q(V_\infty - V_{\text{mid}}) = (1\times 10^{-6})(0-3.6\times 10^5) = \boxed{-0.36\,\text{J}}\). The field does 0.36 J of work on the test charge.
Assertion-Reason Questions
Assertion (A): Electric field is always perpendicular to an equipotential surface.
Reason (R): Any component of \(\vec E\) along the surface would do non-zero work when a charge moves along it.
Answer: A. A tangential field would change \(V\) along the surface, contradicting "equipotential".
Assertion (A): The potential at the equatorial point of a short dipole is zero, but the field there is not zero.
Reason (R): \(V\) is a scalar sum while \(\vec E\) is a vector sum.
Answer: A. Scalar sum vanishes; vector components along the axis add up.
Assertion (A): The potential energy of two like charges is positive.
Reason (R): Work must be done by an external agent to bring like charges close together against their mutual repulsion.
Answer: A. Positive work done is stored as positive PE.
Frequently Asked Questions - Potential and Energy
What is the main concept covered in Potential and Energy?
In NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance), "Potential and Energy" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Potential and Energy useful in real-life applications?
Real-life applications of "Potential and Energy" from NCERT Class 12 Physics Chapter 2 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Potential and Energy?
Key formulas in "Potential and Energy" (NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 2?
NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance) is structured so each part builds on the previous one. "Potential and Energy" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Potential and Energy?
CBSE board questions from "Potential and Energy" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Potential and Energy" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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