Electrostatic potentialWork per unit charge from ∞
VoltSI unit = J/C
Potential differenceV₁ − V₂
Electric dipole±q at small sep. 2a
Dipole momentp = 2aq
Equipotential surfaceV constant; ⊥ to E
Potential energyU = qV
Conservative forcePath-independent work
ConductorFree electrons; E=0 inside
EarthingConnect to V = 0 reference
Electrostatic shieldingCavity has E = 0
Faraday cageHollow conductor as shield
DielectricInsulator that polarises
Polar moleculePermanent dipole
Non-polar moleculeInduced dipole only
Polarisation PDipole moment per volume
Dielectric constant KE₀/Eᵢₙ
CapacitorTwo conductors + insulator
CapacitanceC = Q/V
FaradSI unit, 1 F = 1 C/V
Parallel plate capacitorC = ε₀A/d
Series combination1/Cₑq = Σ 1/Cᵢ
Parallel combinationCₑq = Σ Cᵢ
Energy densityu = ½ε₀E²
Induced chargeBound charges on surface
Permittivity ε₀8.854 × 10⁻¹² F/m
NCERT Exercises — Full Solutions
Exercise 2.1
Two charges \(q_1 = 5\times 10^{-8}\) C and \(q_2 = -3\times 10^{-8}\) C are 16 cm apart. At what point on the line joining them is the potential zero? Take V = 0 at infinity.
Let V = 0 at a point at distance \(x\) cm from \(q_1\) (along the line toward \(q_2\)). Then distance from \(q_2\) is \((16-x)\) cm.
\[\frac{kq_1}{x}+\frac{kq_2}{16-x}=0 \Rightarrow \frac{5}{x}=\frac{3}{16-x}\]
\(5(16-x)=3x \Rightarrow x = 10\) cm from \(q_1\).
Checking the outside region (beyond \(q_2\)): let distance from \(q_1\) be \(y\), from \(q_2\) be \(y-16\). \(\frac{5}{y}=\frac{3}{y-16}\Rightarrow y=40\) cm from \(q_1\).
\(\boxed{\text{Zero at 10 cm (between) and at 40 cm (beyond } q_2\text{) from } q_1.}\)
Exercise 2.2
A regular hexagon of side 10 cm has a charge \(5\,\mu\)C at each vertex. Find the potential at the centre.
The centre of a regular hexagon is equidistant (\(r = \text{side} = 0.10\) m) from each vertex.
\[V = 6\times\frac{kq}{r}=6\times\frac{(9\times 10^9)(5\times 10^{-6})}{0.10}=\boxed{2.7\times 10^{6}\,\text{V}}\]
Exercise 2.3
Two charges \(+1.5\,\mu\)C and \(+2.5\,\mu\)C are 30 cm apart. (a) Identify an equipotential surface. (b) What is the direction of the electric field at every point on the equipotential?
(a) Because the two charges are unequal and of the same sign, no plane is an equipotential. However, the perpendicular-bisector plane is not an equipotential here (symmetry is broken). Equipotentials are curved closed surfaces surrounding both charges, bulging toward the smaller charge. A general equipotential can be sketched by joining points of equal total V computed from the superposition \(V = k(1.5/r_1 + 2.5/r_2)\).
(b) At every point on an equipotential, \(\vec E\) is normal to the surface and points in the direction of decreasing V (outward, away from both charges in this case).
Exercise 2.4
A spherical conductor of radius 12 cm has a charge of \(1.6\times 10^{-7}\) C distributed uniformly on its surface. Find the field (a) inside, (b) just outside, (c) at 18 cm from the centre.
A parallel-plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance is halved and the gap is filled by a substance of dielectric constant 6?
Three capacitors of 9 pF each are connected in series. (a) What is the equivalent capacitance? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
(a) \(\dfrac{1}{C_{\text{eq}}} = \dfrac{3}{9}\Rightarrow C_{\text{eq}} = \boxed{3\,\text{pF}}\).
(b) Same charge on each: \(Q = C_{\text{eq}}V = 3\times 10^{-12}\times 120 = 3.6\times 10^{-10}\) C. Voltage across each: \(V_i = Q/C_i = 3.6\times 10^{-10}/9\times 10^{-12} = \boxed{40\,\text{V}}\). (Sum 120 V ✓)
Exercise 2.7
Three capacitors of capacitance 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance? (b) Find the charge on each if the combination is connected to a 100 V supply.
(a) \(C_{\text{eq}} = 2+3+4 = \boxed{9\,\text{pF}}\).
(b) Each has V = 100 V: \(Q_1 = 200\) pC, \(Q_2 = 300\) pC, \(Q_3 = 400\) pC. Total \(Q = 900\) pC.
Exercise 2.8
A parallel-plate capacitor with plates of area \(6\times 10^{-3}\) m² and separation 3 mm has air between the plates. (a) Find its capacitance. (b) What charge appears on each plate when it is connected to a 100 V supply? (c) What would be the capacitance and the charge if a 3 mm mica sheet (K = 6) were inserted while the battery remained connected?
(a) \(C = \varepsilon_0 A/d = (8.854\times 10^{-12})(6\times 10^{-3})/3\times 10^{-3} = \boxed{1.77\times 10^{-11}\,\text{F} = 17.7\,\text{pF}}\).
(b) \(Q = CV = 1.77\times 10^{-11}\times 100 = \boxed{1.77\times 10^{-9}\,\text{C}}\).
(c) With mica: \(C' = 6\times 17.7 = 106.2\) pF; voltage still 100 V (battery connected), so \(Q' = 6\times 1.77 = \boxed{1.062\times 10^{-8}\,\text{C}}\). (Extra charge flows from the battery.)
Exercise 2.9
Repeat Exercise 2.8 (c) assuming the battery is disconnected before the mica sheet is introduced.
Now Q is fixed at 1.77 × 10⁻⁹ C. C becomes \(K C_0 = 106.2\) pF. New voltage \(V = Q/C = 1.77\times 10^{-9}/1.062\times 10^{-10} = \boxed{16.67\,\text{V}}\). (Reduced by factor K = 6.)
Exercise 2.10
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored?
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected and connected to another uncharged 600 pF capacitor. How much energy is lost?
Initial \(U_i = \tfrac12(600\times 10^{-12})(200)^2 = 1.2\times 10^{-5}\) J.
After sharing, \(C_{\text{tot}} = 1200\) pF with Q = 1.2 × 10⁻⁷ C, so \(V_f = 100\) V. \(U_f = \tfrac12(1200\times 10^{-12})(100)^2 = 6\times 10^{-6}\) J.
\[\Delta U = U_i - U_f = \boxed{6\times 10^{-6}\,\text{J lost}}\]
Exercise 2.12
A charge of \(8\,\text{mC}\) is located at the origin. Calculate the work done in carrying \(-2\times 10^{-9}\) C from a point P (0, 0, 3 cm) to Q (0, 4 cm, 0).
A cube of side \(b\) has a charge \(q\) at each of its eight corners. Find the potential at the centre of the cube.
Diagonal of cube = \(b\sqrt 3\); distance from centre to each corner = \(b\sqrt 3/2\).
\[V = 8\times\frac{kq}{b\sqrt 3/2} = \frac{16\,kq}{b\sqrt 3}=\boxed{\frac{16\,q}{4\pi\varepsilon_0\,\sqrt 3\,b}}\]
Exercise 2.14
Two tiny spheres carry charges \(1.5\,\mu\)C and \(2.5\,\mu\)C, located 30 cm apart. Find the electric potential at the midpoint of the line joining the two charges.
Midpoint is 15 cm from each.
\[V = \frac{k(q_1+q_2)}{r} = \frac{(9\times 10^9)(4\times 10^{-6})}{0.15}=\boxed{2.4\times 10^{5}\,\text{V}}\]
Exercise 2.15
A spherical conducting shell of inner radius \(r_1\) and outer radius \(r_2\) has a charge Q. (a) Is the charge given to the shell on the inner or outer surface? (b) A point charge \(q\) is placed at the centre of the cavity — what are the surface charges now?
(a) Entirely on the outer surface (no field in conductor, nothing to terminate on the inner surface).
(b) With \(q\) at the centre: an induced charge \(-q\) appears on the inner surface, and the outer surface carries \(Q + q\).
Exercise 2.16
Show that the normal component of electric field has a discontinuity from one side of a charged surface to another given by \((\vec E_2 - \vec E_1)\cdot\hat n = \sigma/\varepsilon_0\), where \(\sigma\) is the surface charge density. Hence show that just outside a conductor \(E = \sigma/\varepsilon_0\).
Construct a Gaussian pillbox of area \(dA\) straddling the surface, thickness → 0. By Gauss's law, net flux = \(\sigma\,dA/\varepsilon_0\); only the two normal faces contribute, giving \((E_{2n}-E_{1n})dA = \sigma dA/\varepsilon_0\). Thus \((\vec E_2 - \vec E_1)\cdot\hat n=\sigma/\varepsilon_0\). For a conductor, \(\vec E_1 = 0\) inside, so \(E_{\text{outside}} = \sigma/\varepsilon_0\).
Exercise 2.17
Show that the tangential component of E is continuous across a surface charge. Why does this suggest that field lines run normal to a conductor?
Take a rectangular Amperian loop straddling the surface, width \(\ell\) parallel to surface, height → 0. Since \(\oint\vec E\cdot d\vec l = 0\) (electrostatic field is conservative), \((E_{t,2}-E_{t,1})\ell=0\Rightarrow E_{t,2}=E_{t,1}\). Inside a conductor \(E_t = 0\); therefore just outside \(E_t = 0\) too. Only the normal component survives — field lines exit perpendicularly.
Exercise 2.18
A long charged cylinder of linear charge density \(\lambda\) is surrounded by a coaxial hollow conducting cylinder. What is the electric field in the space between the cylinders?
Apply Gauss's law to a coaxial cylinder of radius \(r\) between the inner wire and outer cylinder:
\[E(2\pi r L) = \lambda L/\varepsilon_0 \Rightarrow E = \frac{\lambda}{2\pi\varepsilon_0 r}\]
directed radially outward (if \(\lambda>0\)).
Exercise 2.19
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å. Estimate the potential energy of the system (in eV), taking the zero of potential energy at infinite separation.
If one of two electrons in H₂ is removed, we get H₂⁺ ion. In the ground state the two protons are 1.06 Å apart and the electron is 0.53 Å from each. Find the PE of the system (take V = 0 at infinity).
Two charged conducting spheres of radii \(a\) and \(b\) are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres?
Connected ⇒ same potential: \(kQ_a/a = kQ_b/b\Rightarrow Q_a/Q_b = a/b\). Surface field \(E = kQ/R^2\), so
\[\frac{E_a}{E_b}=\frac{Q_a/a^2}{Q_b/b^2}=\frac{a/b\cdot b^2}{a^2}=\boxed{\frac{b}{a}}\]
The smaller sphere has the larger surface field — the "corona effect" behind lightning rods.
Exercise 2.22
Two capacitors of capacitances 2 μF and 3 μF are charged to common potential 100 V and connected in parallel with similar polarity (+ to +). Find the common voltage and energy lost.
Already at the same potential, so no charge flows — common voltage remains 100 V, no energy is lost. If instead they are connected with opposite polarity:
\(Q_1 = 2\times 10^{-6}\times 100 = 2\times 10^{-4}\) C (+); \(Q_2 = 3\times 10^{-6}\times 100 = 3\times 10^{-4}\) C (−). Net \(Q = 10^{-4}\) C.
\(V_f = 10^{-4}/5\times 10^{-6} = 20\) V. \(U_i = \tfrac12(2+3)\times 10^{-6}\times 100^2 = 0.025\) J. \(U_f = \tfrac12(5\times 10^{-6})(20)^2 = 10^{-3}\) J. \(\Delta U = \boxed{0.024\,\text{J lost}}\).
Exercise 2.23
A capacitor of 4 μF is charged to 200 V, disconnected, then a dielectric slab of K = 4 is inserted filling the gap. Find (a) new capacitance, (b) new voltage, (c) energy before and after. Explain any discrepancy.
(a) \(C' = 4\times 4 = 16\,\mu\)F.
(b) Q is fixed at \(8\times 10^{-4}\) C; \(V' = Q/C' = 8\times 10^{-4}/16\times 10^{-6}=50\) V.
(c) \(U_i = \tfrac12(4\times 10^{-6})(200)^2 = 0.08\) J; \(U_f = \tfrac12(16\times 10^{-6})(50)^2 = 0.02\) J.
Energy drops by a factor of 4; the dielectric is pulled into the capacitor, and the agent (or bound-charge system) does negative work — the lost field energy is spent on polarisation plus a small amount of work done on whatever holds the slab.
Exercise 2.24
Compute the equivalent capacitance of the network below: three capacitors \(C_1 = 100\) pF, \(C_2 = 200\) pF, \(C_3 = 200\) pF are connected such that \(C_2\) and \(C_3\) are in parallel, and their combination is in series with \(C_1\) across 300 V. Find the charge on each.
\(C_{23} = 200 + 200 = 400\) pF. In series with \(C_1\):
\[\frac{1}{C_{\text{eq}}} = \frac{1}{100}+\frac{1}{400} = \frac{5}{400}\Rightarrow C_{\text{eq}} = 80\,\text{pF}\]
Total charge \(Q = C_{\text{eq}}V = 80\times 10^{-12}\times 300 = 2.4\times 10^{-8}\) C.
Since \(C_1\) and \(C_{23}\) are in series, \(Q_1 = Q_{23} = 2.4\times 10^{-8}\) C.
Voltages: \(V_1 = Q/C_1 = 240\) V; \(V_{23} = 60\) V.
Charges on parallel caps: \(Q_2 = 200\times 10^{-12}\times 60 = 1.2\times 10^{-8}\) C; \(Q_3 = 1.2\times 10^{-8}\) C.
\[\boxed{C_{\text{eq}} = 80\,\text{pF};\;Q_1 = 24\,\text{nC};\;Q_2 = Q_3 = 12\,\text{nC}}\]
Frequently Asked Questions - NCERT Exercises and Solutions: Electrostatic Potential and Capacitance
What are the key NCERT exercise types in Chapter 2 Electrostatic Potential and Capacitance?
NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Electrostatic Potential and Capacitance?
For numerical problems in NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 2?
From NCERT Class 12 Physics Chapter 2 (Electrostatic Potential and Capacitance), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 2 Electrostatic Potential and Capacitance problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 2 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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