Chapter 2 of 14

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CBSE Class 12 Physics★ Unit I · Electrostatics, CBSE states no chapter-wise split

Electrostatic Potential and Capacitance

Potential as a scalar, potential due to point charges and dipoles, equipotential surfaces, potential energy, conductors and dielectrics, and capacitors in series and parallel with their stored energy. The Additional Exercises block has been removed, leaving 11 questions. CBSE Class 12 Physics Chapter 2.

⏱ 12 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Compute the potential due to a point charge, a dipole and a system of charges by algebraic addition
2Identify equipotential surfaces and explain why the field is always perpendicular to them
3Calculate the potential energy of a system of charges by summing over pairs
4State the electrostatic properties of a conductor and explain electrostatic shielding
5Explain polarisation and how a dielectric raises capacitance by the factor K
6Combine capacitors in series and parallel and compute the energy stored

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Why this chapter matters

Potential is a scalar, which makes it far easier to work with than the field of the previous chapter, and capacitance is the bridge from electrostatics into circuits. The series and parallel rules and the energy formulas reappear directly in Current Electricity and Alternating Current.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Electric Charges and Fields (Class 12 Ch 1): potential is defined through work done against the electric field, so Coulomb's law and the field concept must be secure

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Work, energy and power (Class 11): potential is work per unit charge, and potential energy is the work done in assembling a configuration

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Scalar addition versus vector addition (Class 11): the central simplification here is that potentials add algebraically while fields add vectorially

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Combinations and counting (Class 11): the potential energy of n charges has n(n-1)/2 pair terms, which is where most errors arise

Electrostatic Potential and Capacitance

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as it has from all 14 chapters of the current Class 12 Physics book. A full-text search of every chapter PDF returns zero occurrences of the phrase, and the questions here run contiguously from 2.1 to 2.11 with no gaps.

That leaves 11 questions for a 36-page chapter — the thinnest question-to-content ratio in the Electrostatics unit, and well short of what the chapter's fifteen sections actually cover. Several major topics carry no exercise question at all: potential due to a dipole (2.4), potential energy in an external field (2.8), and dielectrics and polarisation (2.10) are each taught in full and then never tested here.

The old stub had no solutions file, and duplicated meta-driven sections as ## headings inside the page body.

Textbook section	Topic
2.2 to 2.5	Electrostatic potential; potential due to a point charge, a dipole and a system of charges
2.6	Equipotential surfaces
2.7 to 2.8	Potential energy of a system of charges, and in an external field
2.9 to 2.10	Electrostatics of conductors; dielectrics and polarisation
2.11 to 2.13	Capacitors and capacitance; the parallel plate capacitor; effect of a dielectric
2.14 to 2.15	Combination of capacitors; energy stored in a capacitor

The same micro-symbol trap as Chapter 1 applies. Under text extraction " $5 μ$ C" appears as "5 mC", a factor of a thousand. Exercise 2.2's hexagon gives the standard $2.7 \times 1 0^{6}$ V with $μ$ C; reading mC would give $2.7 \times 1 0^{9}$ V. All values here were checked against the printed page.

2. Electrostatic Potential (Textbook 2.2 to 2.5)

The electrostatic potential at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point:

$V = \frac{W}{q _{0}}$

Its unit is the volt, $1$ V $= 1$ J/C. Potential is a scalar, and this single fact is what makes the whole chapter easier than the last one — contributions add algebraically, with no vector resolution.

Potential due to a point charge (2.3):

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

Note it falls off as $1/ r$ , not $1/ r^{2}$ as the field does. The sign of $q$ carries straight through, so a negative charge gives a negative potential.

Potential due to a system of charges (2.5) is the plain algebraic sum:

$V = k (\frac{q _{1}}{r _{1}} + \frac{q _{2}}{r _{2}} + \dots)$

Exercise 2.2 exploits this directly: six equal charges at the vertices of a regular hexagon each sit a distance equal to the side from the centre, so the potentials simply add to $6 k q / r$ . The field at that point is zero by symmetry, while the potential is not — a contrast worth holding onto.

Potential due to a dipole (2.4):

$V = \frac{k p c o s θ}{r ^{2}}$

This falls off as $1/ r^{2}$ , faster than a point charge's $1/ r$ . On the equatorial line $θ = 9 0^{\circ}$ , so $V = 0$ everywhere on it, even though the field there is not zero.

Relation between field and potential. The field is the negative gradient of the potential:

$E = - \frac{d V}{d r}$

The minus sign says the field points from high potential towards low potential.

3. Equipotential Surfaces (Textbook 2.6)

An equipotential surface is one on which the potential has the same value at every point.

The field is always perpendicular to an equipotential surface. If it had any component along the surface, work would be done in moving a charge between two points at the same potential — and by definition that work is zero. So no tangential component can exist.

Two consequences follow immediately: no work is done in moving a charge along an equipotential surface, and two equipotential surfaces can never intersect, since that would give two potentials at one point.

Charge configuration	Equipotential surfaces
Single point charge	Concentric spheres centred on the charge
Uniform field	Planes perpendicular to the field
Two equal and opposite charges	The perpendicular bisector plane is the $V = 0$ surface

That last row is Exercise 2.3: every point on the perpendicular bisector of the line joining $+ q$ and $- q$ is equidistant from both, so the two potentials cancel exactly and $V = 0$ .

Where surfaces are closely spaced, the potential changes rapidly with distance, so the field is strong there.

4. Potential Energy of a System of Charges (Textbook 2.7 to 2.8)

The potential energy of a system is the work done in assembling it, bringing each charge from infinity against the field of those already placed.

For two charges:

$U = \frac{k q _{1} q _{2}}{r _{12}}$

For three charges, add the work for each pair — there are three pairs, not three terms:

$U = k (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{1} q _{3}}{r _{13}} + \frac{q _{2} q _{3}}{r _{23}})$

Counting pairs rather than charges is the usual difficulty. For $n$ charges there are $n (n - 1) /2$ pairs.

The signs matter: $U$ is positive for like charges (work must be done to push them together) and negative for unlike charges (they attract, so the system releases energy).

Potential energy in an external field (2.8). For a single charge $q$ at a point where the external potential is $V$ , the energy is $U = q V$ .

For a dipole in a uniform external field:

$U = - pE cos θ = - p \cdot E$

This is least at $θ = 0$ , where the dipole is aligned with the field, which is therefore the position of stable equilibrium. At $θ = 18 0^{\circ}$ the energy is greatest and the equilibrium is unstable.

5. Conductors and Dielectrics (Textbook 2.9 to 2.10)

Electrostatics of conductors (2.9). In electrostatic equilibrium a conductor obeys several rules, all following from the fact that free charges move until no force acts on them:

The field inside the conductor is zero, since any field would drive the free electrons until it was cancelled.
Any excess charge resides entirely on the surface, because charges repel and move as far apart as possible.
The field just outside is perpendicular to the surface; a tangential component would drive surface currents.
The whole conductor is an equipotential, including its interior and surface, since $E = 0$ inside means no potential difference.

The zero interior field is the basis of electrostatic shielding — a metal enclosure protects whatever is inside it from external fields.

Dielectrics and polarisation (2.10). A dielectric is an insulator with no free charges. Placed in an external field, its molecules develop or align dipole moments, which is polarisation.

The aligned dipoles produce their own field opposing the applied one, so the net field inside the dielectric is reduced:

$E_{net} = \frac{E _{0}}{K}$

where $K$ is the dielectric constant. Since the field is weakened but the charge is unchanged, the potential difference falls and the capacitance rises by the same factor $K$ — which is exactly why dielectrics are used in capacitors.

6. Capacitors, Combinations and Stored Energy (Textbook 2.11 to 2.15)

A capacitor stores charge. For any capacitor the charge is proportional to the potential difference:

$Q = C V, C = \frac{Q}{V}$

Capacitance is measured in farads, $1$ F $= 1$ C/V. The farad is enormous, so practical values are in $μ$ F or pF.

Parallel plate capacitor (2.12 to 2.13):

$C = \frac{ε _{0} A}{d}, and with a dielectric C = \frac{K ε _{0} A}{d}$

So capacitance rises with plate area, falls with separation, and is multiplied by $K$ when a dielectric fills the gap. Exercise 2.5 combines two of these: halving $d$ doubles $C$ and a dielectric of $K = 6$ multiplies it by 6, giving a factor of 12 overall.

Combinations (2.14). The two cases behave oppositely, and confusing them is the commonest error in the chapter.

	Series	Parallel
Formula	$\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots$	$C_{p} = C_{1} + C_{2} + \dots$
Same for all	Charge $Q$	Voltage $V$
Result	Smaller than the smallest	Larger than the largest

Note this is the reverse of how resistors combine, which is a frequent source of confusion.

Energy stored (2.15). The three equivalent forms follow from one another using $Q = C V$ :

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

Choose whichever form matches the quantities you are given. The voltage is squared, so halving the supply voltage quarters the stored energy.

Energy is not conserved when capacitors are connected together. In Exercise 2.11 a charged capacitor is joined to an identical uncharged one. The charge is conserved, but sharing it across twice the capacitance halves the voltage, and since $U \propto V^{2}$ at fixed charge, exactly half the energy is lost — dissipated as heat in the connecting wires and as radiation during the transient current.

Summary

Potential is $V = W / q_{0}$ in volts, and it is a scalar — contributions add algebraically with no vector resolution.
Point charge: $V = k q / r$ , falling off as $1/ r$ while the field falls off as $1/ r^{2}$ ; the sign of $q$ carries through.
Dipole: $V = k p cos θ / r^{2}$ , which is zero everywhere on the equatorial line although the field there is not.
The field is the negative gradient of potential, $E = - d V / d r$ , pointing from high to low potential.
The field is always perpendicular to an equipotential surface, no work is done moving along one, and two such surfaces never intersect.
For two equal and opposite charges the perpendicular bisector plane is the $V = 0$ equipotential surface.
Potential energy is summed over pairs: three charges give three terms, and $n$ charges give $n (n - 1) /2$ .
A dipole in an external field has $U = - p \cdot E$ , minimum and stable at $θ = 0$ .
In a conductor at equilibrium: zero interior field, charge entirely on the surface, field perpendicular just outside, and the whole body an equipotential.
A dielectric polarises and opposes the applied field, reducing it to $E_{0} / K$ and raising the capacitance by $K$ .
$Q = C V$ ; parallel plate $C = K ε_{0} A / d$ .
Series: reciprocals add, charge is common, result smaller than the smallest. Parallel: capacitances add, voltage is common, result larger than the largest. This is the reverse of resistors.
Energy stored is $\frac{1}{2} C V^{2} = \frac{1}{2} Q V = Q^{2} /2 C$ ; the voltage is squared.
Joining a charged capacitor to an identical uncharged one conserves charge but loses half the energy as heat and radiation.
The Additional Exercises block has been removed, leaving Exercises 2.1 to 2.11.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Electrostatic potential

V = W/q0, measured in volts, where 1 V = 1 J/C

A scalar, so contributions add algebraically with no vector resolution

Potential due to a point charge

V = k q / r

Falls off as 1/r, not 1/r^2 as the field does; the sign of q carries straight through

Potential due to a system of charges

V = k(q1/r1 + q2/r2 + ...)

A plain algebraic sum — this is why the hexagon in Exercise 2.2 needs no geometry beyond the common distance

Potential due to a dipole

V = k p cos(theta) / r^2

Falls off as 1/r^2; on the equatorial line theta is 90 degrees so V = 0 although the field there is not zero

Field from potential

E = -dV/dr

The minus sign means the field points from high potential towards low potential

Equipotential surfaces

Surfaces of constant V; the field is everywhere perpendicular to them

No work is done moving a charge along one, and two equipotential surfaces can never intersect

Potential energy of two charges

U = k q1 q2 / r12

Positive for like charges and negative for unlike charges

Potential energy of a system

Sum over all PAIRS: for three charges, U = k(q1q2/r12 + q1q3/r13 + q2q3/r23)

n charges give n(n-1)/2 pair terms, not n terms — the commonest slip in this section

Dipole in an external field

U = -p E cos(theta) = -p . E

Minimum at theta = 0, which is stable equilibrium; maximum at 180 degrees, which is unstable

Properties of a conductor at equilibrium

E = 0 inside; charge entirely on the surface; E perpendicular just outside; the whole conductor is an equipotential

All four follow from free charges moving until no force acts on them; the zero interior field gives electrostatic shielding

Effect of a dielectric

E_net = E0 / K, where K is the dielectric constant

Polarised molecules set up a field opposing the applied one, reducing the net field and raising capacitance by K

Definition of capacitance

Q = CV, so C = Q/V, measured in farads where 1 F = 1 C/V

The farad is very large, so practical capacitors are rated in microfarads or picofarads

Parallel plate capacitor

C = epsilon_0 A / d, and with a dielectric C = K epsilon_0 A / d

Rises with plate area, falls with separation, and is multiplied by K when a dielectric fills the gap

Capacitors in series

1/Cs = 1/C1 + 1/C2 + ...; the CHARGE is the same on each

The result is smaller than the smallest capacitor — the reverse of how resistors behave in series

Capacitors in parallel

Cp = C1 + C2 + ...; the VOLTAGE is the same across each

The result is larger than the largest capacitor

Energy stored in a capacitor

U = (1/2)CV^2 = (1/2)QV = Q^2/(2C)

Use whichever form matches the given quantities; the voltage is squared, so halving V quarters the energy

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Adding potentials as vectors, or resolving them into components

✓ Potential is a scalar. Add the contributions algebraically with their signs. Only the field requires vector addition.

WATCH OUT

✗ Using 1/r squared for potential due to a point charge

✓ Potential goes as kq/r; it is the field that goes as kq/r^2. Confusing the two changes the answer by a factor of r.

WATCH OUT

✗ Summing potential energy over charges instead of over pairs

✓ Three charges give three PAIR terms, and n charges give n(n-1)/2. Write out the pairs explicitly before substituting.

WATCH OUT

✗ Swapping the series and parallel formulas for capacitors

✓ Capacitors add directly in parallel and reciprocally in series — the reverse of resistors. In series the charge is common; in parallel the voltage is common.

WATCH OUT

✗ Forgetting which quantity stays fixed when a dielectric is inserted

✓ If the supply remains connected, V is fixed and Q rises by K. If the supply is disconnected, Q is fixed and V falls by K. Exercise 2.9 tests exactly this distinction.

WATCH OUT

✗ Failing to square the voltage in the energy formula

✓ U = (1/2)CV^2. Halving the supply voltage quarters the stored energy rather than halving it.

WATCH OUT

✗ Assuming energy is conserved when two capacitors are connected together

✓ Charge is conserved but energy is not. Joining a charged capacitor to an identical uncharged one loses exactly half the energy as heat in the wires and as radiation.

WATCH OUT

✗ Leaving the plate separation in millimetres

✓ Convert to metres before substituting into C = epsilon_0 A/d. Exercise 2.8 gives 3 mm, which is 3 x 10^-3 m.

WATCH OUT

✗ Assuming the field is zero wherever the potential is zero

✓ They are independent. On a dipole's equatorial line V = 0 but E is not, and at the centre of a charged hexagon E = 0 but V is not.

WATCH OUT

✗ Reading the micro symbol as milli in an extracted PDF

✓ The micro symbol often extracts as a plain 'm'. Exercise 2.2 gives 2.7 x 10^6 V with microcoulomb but an impossible 2.7 x 10^9 V with millicoulomb.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Potential due to charge systems, equipotential surfaces, conductors, and capacitors in series and parallel with their stored energy

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Electrostatic Potential and Capacitance?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•V = W/q0 in volts; potential is a scalar and adds algebraically
•Point charge: V = kq/r, falling as 1/r while the field falls as 1/r^2
•System of charges: V is the plain algebraic sum of the individual potentials
•Dipole: V = k p cos(theta)/r^2, zero on the equatorial line where the field is not zero
•E = -dV/dr, so the field points from high potential to low potential
•The field is always perpendicular to an equipotential surface
•No work is done moving a charge along an equipotential surface
•Two equipotential surfaces can never intersect
•For equal and opposite charges, the perpendicular bisector plane is the V = 0 surface
•Potential energy sums over PAIRS: n charges give n(n-1)/2 terms
•U is positive for like charges and negative for unlike charges
•Dipole in a field: U = -p.E, minimum and stable when aligned with the field
•Conductor at equilibrium: zero interior field, surface charge only, perpendicular exterior field, whole body an equipotential
•Electrostatic shielding follows from the zero field inside a conducting enclosure
•A dielectric polarises and opposes the applied field, giving E_net = E0/K and raising C by K
•Q = CV; parallel plate C = K epsilon_0 A/d
•Series: reciprocals add, charge common, result smaller than the smallest
•Parallel: capacitances add, voltage common, result larger than the largest
•Capacitors combine in the reverse manner to resistors
•Energy stored: U = half CV squared = half QV = Q squared over 2C
•Joining a charged capacitor to an identical uncharged one loses half the energy

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I: Electrostatics, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Electrostatic Potential and Potential Energy	3-4	1	Potential of charge systems, equipotential surfaces, and pairwise potential energy
Conductors and Dielectrics	2-3	1	Properties of a conductor at equilibrium, shielding, and the effect of polarisation on the net field
Capacitors, Combinations and Energy	4-5	1	Parallel plate capacitance, series and parallel combinations, dielectric effects, and stored energy

Prep strategy

Exploit the fact that potential is a scalar — it removes all the vector work that made the previous chapter harder
When a dielectric is inserted, always first ask whether the supply is connected (V fixed) or disconnected (Q fixed)
Write out the pairs explicitly before computing the potential energy of a multi-charge system
Remember that capacitors combine oppositely to resistors, and state which quantity is common in each case
Check whether a question asks for potential or field — the two are zero in different places

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Camera flash units

A capacitor charges slowly from a small battery and then discharges almost instantly through the flash tube, delivering far more power than the battery could supply directly.

Defibrillators

A large capacitor stores energy and releases a controlled pulse through the chest, an application resting entirely on the half CV squared energy formula.

Touchscreens

Capacitive screens detect a finger by the change it makes to the local capacitance, since the human body acts as a dielectric and alters the stored charge.

Electrostatic shielding of electronics

Sensitive circuits are enclosed in metal cases because the field inside a conductor is zero, protecting them from external interference.

Power factor correction

Capacitor banks are installed in electrical substations to store and return energy each cycle, improving the efficiency of power transmission.

Computer memory

Each cell of dynamic RAM is a tiny capacitor whose charge represents a bit, refreshed thousands of times a second as the charge leaks away.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

State explicitly that potential is a scalar when adding contributions, which earns the method mark

For dielectric questions, write down first whether V or Q is being held constant, then derive the rest

List the pairs before computing potential energy for three or more charges

Convert plate separations from millimetres to metres at the very start of any capacitance calculation

Name which quantity is common when combining capacitors: charge in series, voltage in parallel

Show the intermediate value of the equivalent capacitance before computing charges or energies

When asked for energy lost, compute the initial and final energies separately and subtract, rather than quoting a memorised fraction

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Deriving the potential of a dipole at an arbitrary point by binomial expansion, and showing the 1/r^2 dependence emerges from the cancellation of the leading terms

STRETCH

The energy density of an electric field, u = half epsilon_0 E squared, which recovers the capacitor energy formula by integration over the volume between the plates

STRETCH

Capacitance of an isolated sphere and of concentric spherical and cylindrical shells, obtained from the potential difference between the conductors

STRETCH

Force between the plates of a charged capacitor, derived by differentiating the stored energy with respect to separation

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainDielectric inserted with the supply connected or disconnectedIdentifying which quantity is held fixed

A parallel plate capacitor of capacitance $17.7$ pF is connected to a 100 V supply. A mica sheet of dielectric constant 6 is inserted (a) with the supply connected, (b) after it is disconnected. Find the change in charge and voltage in each case.

Stuck? Show the approach

Decide in each case whether the battery fixes V or the isolation fixes Q, then let the rest follow from Q = CV.

Show the full solution

In both cases the dielectric raises the capacitance to $C^{'} = 6 \times 17.7 = 106$ pF. (a) With the supply connected the battery holds $V = 100$ V, so $Q^{'} = C^{'} V$ rises sixfold from $1.77 \times 1 0^{- 9}$ C to $1.06 \times 1 0^{- 8}$ C, the extra charge coming from the battery. (b) Once disconnected the charge is trapped at $1.77 \times 1 0^{- 9}$ C, so $V^{'} = Q / C^{'}$ falls to one sixth, giving $100/6 \approx 16.7$ V.

Answer: (a) V stays 100 V, Q rises to 1.06 x 10^-8 C (b) Q stays 1.77 x 10^-9 C, V falls to about 16.7 V

The trap

Applying the same conclusion to both parts. The entire question turns on whether the battery is present to hold the voltage constant.

JEE MainEnergy lost on connecting capacitorsCharge conserved but energy not

A 600 pF capacitor charged by a 200 V supply is disconnected and then connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost?

Stuck? Show the approach

Conserve charge to find the common final voltage, then compare initial and final stored energies.

Show the full solution

Initially $Q = 600 \times 1 0^{- 12} \times 200 = 1.2 \times 1 0^{- 7}$ C and $U_{i} = \frac{1}{2} C V^{2} = 1.2 \times 1 0^{- 5}$ J. After connection the charge is unchanged but the total capacitance is $1200$ pF, so $V_{f} = \frac{1.2 \times 1 0 ^{- 7}}{1200 \times 1 0 ^{- 12}} = 100$ V. Then $U_{f} = \frac{1}{2} \times 1200 \times 1 0^{- 12} \times 10 0^{2} = 6 \times 1 0^{- 6}$ J. The loss is $6 \times 1 0^{- 6}$ J, exactly half.

Answer: 6 x 10^-6 J, half the initial energy

The trap

Assuming energy is conserved because the system is isolated. Charge is conserved; the missing energy is dissipated as heat in the connecting wires and as radiation during the transient current.

JEE AdvancedZero potential versus zero fieldDistinguishing two independent conditions

Two charges $5 \times 1 0^{- 8}$ C and $- 3 \times 1 0^{- 8}$ C are 16 cm apart. Find all points on the line joining them where the potential is zero.

Stuck? Show the approach

Set the algebraic sum of the two potentials to zero and consider each region of the line separately.

Show the full solution

Setting $\frac{5}{r _{1}} = \frac{3}{r _{2}}$ : between the charges, with $r_{1} = x$ and $r_{2} = 16 - x$ , we get $5 (16 - x) = 3 x$ , so $x = 10$ cm. Beyond the negative charge, with $r_{2} = x - 16$ , we get $5 (x - 16) = 3 x$ , so $x = 40$ cm. There is no such point beyond the positive charge, since there it is both larger in magnitude and nearer.

Answer: 10 cm and 40 cm from the positive charge, both measured along the line towards the negative charge

The trap

Stopping after finding the point between the charges. Because potential is a signed scalar, an external point also satisfies the condition, and both must be reported.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because potential is a scalar and the field is a vector. To find the field of several charges you must resolve each contribution into components and add them vectorially. To find the potential you simply add the numbers with their signs. Exercise 2.2 shows the contrast: six charges at the vertices of a hexagon give a field of zero at the centre by symmetry, but a potential of 2.7 x 10^6 V obtained by straightforward addition.

Yes, both happen and they are independent conditions. On the equatorial line of a dipole the potential is zero everywhere but the field is not. At the centre of a regular hexagon with equal charges at its vertices the field is zero by symmetry but the potential is large. Never infer one from the other.

Adding capacitors in parallel increases the total plate area available for storing charge, so the capacitance grows. Adding them in series effectively increases the total plate separation, so the capacitance falls and the reciprocals add. For resistors the reasoning runs the other way, since a longer path means more resistance. The safest habit is to remember which quantity is common: charge in series, voltage in parallel.

It depends entirely on whether the supply is still connected. If the battery remains connected it holds the voltage fixed, so the charge increases by the factor K, with the extra charge coming from the battery. If the supply has been disconnected the charge is trapped and cannot change, so the voltage falls to one Kth. Exercise 2.9 asks for both cases side by side.

It is dissipated as heat in the resistance of the connecting wires, and a small amount is radiated as electromagnetic waves during the transient current that flows while the charge redistributes. Charge is strictly conserved throughout, but energy is not, because the process is not reversible. When the two capacitors are equal, exactly half the original energy is lost.

If the field had any component along the surface, that component would do work on a charge moved along it. But by definition all points on an equipotential surface are at the same potential, so the work done between any two of them is zero. The only way to satisfy both is for the tangential component to vanish, leaving the field entirely perpendicular.

A conductor contains free electrons. If any field existed inside, those electrons would experience a force and move, and they would keep moving until their redistribution produced an opposing field that exactly cancelled the original. Electrostatic equilibrium is by definition the state in which this movement has stopped, which requires the net internal field to be zero. This is the basis of electrostatic shielding.

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Current Electricity

Drift velocity, Ohm's law and resistivity, the temperature dependence of resistance, internal resistance and terminal voltage, and Kirchhoff's rules applied to networks including the Wheatstone bridge. The Additional Exercises block has been removed, leaving 9 questions. CBSE Class 12 Physics Chapter 3.

Moving Charges and Magnetism

The Lorentz force, circular and helical motion in a magnetic field, the Biot-Savart law, Ampere's circuital law, the solenoid, the force between parallel currents that defines the ampere, and the torque on a current loop. Two topics the chapter's own text still promises — the cyclotron and the toroid — have been removed from it. CBSE Class 12 Physics Chapter 4.

Magnetism and Matter

The bar magnet and its field lines, the magnet as an equivalent solenoid, torque and potential energy of a magnetic dipole, Gauss's law for magnetism, magnetisation and magnetic intensity, and the classification of materials as diamagnetic, paramagnetic or ferromagnetic. The Earth's magnetism, permanent magnets and the hysteresis loop have all been removed. CBSE Class 12 Physics Chapter 5.

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