Electrostatic Potential and Capacitance — NEET notes

Electrostatic Potential

Electrostatic potential at a point is the work done per unit positive test charge in bringing it from infinity to that point. It tells you 'how high' a charge sits on the electric energy landscape — just like altitude tells you about gravitational energy.

V = W q_0 = 1 4 _0 Q r

Electrostatic Potential

Potential assigns a number (scalar) to every point in space. It tells you how much energy a unit positive charge would have at that point.

The electrostatic potential V at a point P is defined as the work done by an external force to bring a unit positive test charge from infinity to that point, without acceleration.

V( r ) = W_ P q_0 (NCERT Eq. 2.1)

• SI unit of potential → Volt (V) = 1 J/C
• Potential is a scalar quantity — no direction, just a number at each point
• Potential at infinity is taken as zero (reference point)
• Positive charges create positive potential around them; negative charges create negative potential
• Work is done by external force AGAINST the electric field

Potential Due to a Point Charge

The potential due to a single point charge falls off as 1/r — slower than the electric field which falls as 1/r^2 . This difference matters.

For a point charge Q placed at the origin, the potential at distance r is obtained by integrating the work done against the electric field from infinity to r .

V = 1 4 _0 Q r (NCERT Eq. 2.4)

• Potential varies as 1/r (NOT 1/r^2 like the electric field)
• For Q > 0 : V > 0 everywhere (decreases as you move away)
• For Q < 0 : V < 0 everywhere (increases toward zero as you move away)
• At r = 0 : potential is theoretically infinite — a point charge is an idealization
• Potential is additive — for multiple charges, just add up individual potentials (no vector addition needed!)

Quick recall

A charge of +4 is placed at the origin. What is the potential at r = 0.2 m ?

V = 9 10^9 4 10^ -6 0.2 = 1.8 10^5 V = 180 kV

Potential Difference

It's not absolute potential but the potential DIFFERENCE between two points that determines how charges move and how much work is done.

The potential difference between points A and B is the work done per unit positive charge in moving from B to A. Charges flow from higher potential to lower potential — the potential difference is what drives them.

V_A - V_B = W_ B A q_0 (NCERT Eq. 2.2)

• Potential difference is also called voltage in circuits
• 1 Volt = 1 Joule per Coulomb of work done
• Positive charges move from high V to low V (like rolling downhill)
• Negative charges move from low V to high V (they 'roll uphill' in the V-landscape)
• If V_A = V_B , no work is done in moving charge from A to B — this defines an equipotential surface

Potential Due to Systems

When multiple charges are present, potentials simply add up as scalars. Dipoles create characteristic potential patterns, and equipotential surfaces give us a visual map of how potential varies in space.

V = 1 4 _0 _i q_i r_i

Potential Due to a System of Charges

The total potential at a point is the algebraic sum of potentials due to each individual charge. No vector gymnastics needed.

For N point charges q_1, q_2, , q_N at distances r_1, r_2, , r_N from point P, the total potential is simply the algebraic sum of individual potentials.

V = 1 4 _0 _ i=1 ^ N q_i r_i (NCERT Eq. 2.6)

• Potential obeys the superposition principle — each charge contributes independently
• Include signs: +q gives +V , -q gives -V at any point
• If V = 0 at a point, it does NOT mean E = 0 there (classic NEET trap!)
• For continuous charge distributions, replace the sum with an integral: V = 1 4 _0 dq r

Quick recall

Charges +2 and -2 are 10 cm apart. What's the potential at the midpoint?

At the midpoint, r = 5 cm for both. V = kQ/r + k(-Q)/r = 0 . The potentials cancel because of equal magnitudes and opposite signs. But E 0 here — both fields point in the same direction!

Potential Due to an Electric Dipole

A dipole's potential falls faster ( 1/r^2 ) than a single charge ( 1/r ) because the two charges partially cancel each other's potential at distant points.

For an electric dipole with moment p = q 2a p , the potential at a point P at distance r (where r 2a ) and angle from the dipole axis is:

V = 1 4 _0 p r^2 (NCERT Eq. 2.10)

• Axial point ( = 0° ): V = p 4 _0 r^2 — maximum positive value
• Axial point ( = 180° ): V = - p 4 _0 r^2 — maximum negative value
• Equatorial point ( = 90° ): V = 0 — the entire equatorial plane is at zero potential
• Dipole potential falls as 1/r^2 (faster than monopole's 1/r )
• This formula is valid only for r 2a (far-field approximation)

Equipotential Surfaces

An equipotential surface connects all points at the same potential. No work is done moving along such a surface, and the electric field is always perpendicular to it.

• Work done in moving a charge along an equipotential surface = zero (since V = 0 )
• E is always perpendicular to the equipotential surface at every point
• Equipotential surfaces never intersect (a point can't have two different potentials)
• Closer spacing of equipotential surfaces → stronger electric field
• For a point charge: equipotential surfaces are concentric spheres
• For a uniform field: equipotential surfaces are parallel planes perpendicular to E

The direction of E is the direction in which potential decreases most rapidly. This is why E must be perpendicular to equipotential surfaces — any component along the surface would mean potential is changing along the surface, contradicting the definition.

Quick recall

Can two equipotential surfaces at different potentials ever intersect?

No! If they intersected, a point on the intersection would have two different potential values simultaneously — which is impossible. This is exactly like contour lines on a map — the 100 m and 200 m lines never cross.

Relation Between Electric Field and Potential

Electric field and potential are two faces of the same coin. The field is the negative gradient of potential — it points in the direction of steepest potential decrease, like water flowing downhill.

E = - dV dr

Electric Field from Potential

If you know how potential varies in space, you can find the electric field by taking the derivative. The negative sign means E points from high V to low V.

The electric field component in any direction is the negative rate of change of potential in that direction. For a one-dimensional case along the r -axis:

E = - dV dr (NCERT Eq. 2.12)

In three dimensions, the electric field is the negative gradient of potential:

E = - V = - ( V x i + V y j + V z k )

• The negative sign means E points from high V to low V
• SI unit of E from this relation: V/m (equivalent to N/C )
• If V is constant in a region, E = 0 there (inside a conductor, for example)
• If V varies linearly ( V = ax + b ), the field is uniform ( E = -a )
• For a point charge: V = kQ/r , so E = -d(kQ/r)/dr = kQ/r^2 — consistent!

Quick recall

If V = 3x^2 + 4y , what is E ?

E_x = - V/ x = -6x , E_y = - V/ y = -4 , E_z = 0 . So E = -6x i - 4 j V/m .

Potential from Electric Field

If you know the electric field, you can find the potential difference by integrating along any path. The result is path-independent because the electrostatic field is conservative.

The potential difference between two points A and B is obtained by integrating the electric field along ANY path from A to B (path-independence is guaranteed for conservative fields).

V_A - V_B = - _B^A E d l = _A^B E d l (NCERT Eq. 2.11)

• The integral is path-independent — choose any convenient path
• For a uniform field: V_A - V_B = E d where d is the distance along E
• Work done by field on charge q : W = q(V_A - V_B) = -q _A^B E d l
• Along an equipotential surface: E d l = 0 because E d l

Electrostatic Potential Energy

Potential energy is the work stored when you assemble charges from infinity. For two charges it's simple; for a dipole in a field, it depends on orientation — and this is how electric motors and molecular alignment work.

U = 1 4 _0 q_1 q_2 r_ 12

PE of a Two-Charge System

The potential energy of two point charges is the work done to bring one charge from infinity to its position in the field of the other.

The electrostatic potential energy of a system of two point charges q_1 and q_2 separated by distance r_ 12 is the work done in assembling this configuration, starting from all charges at infinity.

U = 1 4 _0 q_1 q_2 r_ 12 (NCERT Eq. 2.14)

• If q_1 and q_2 have the same sign: U > 0 (repulsive, energy was stored by pushing them together)
• If q_1 and q_2 have opposite signs: U < 0 (attractive, energy was released when they came together)
• U = 0 when r (charges infinitely far apart)
• PE depends on the product q_1 q_2 , NOT on which charge was brought first
• Unit: Joule (J) or electron-volt (eV) where 1 eV = 1.6 10^ -19 J

Quick recall

Two charges +3 and -4 are 30 cm apart. Find the PE of the system.

U = 9 10^9 (3 10^ -6 )(-4 10^ -6 ) 0.3 = -0.36 J . Negative PE means the system is bound — you'd need to do 0.36 J of work to separate them to infinity.

PE of a System of Charges

For N charges, the total PE is the sum over all unique pairs. Be careful not to double-count!

For a system of N charges, the total potential energy is the sum of PE of every unique pair. For 3 charges, that's 3 pairs; for 4 charges, 6 pairs; in general, N(N-1) 2 pairs.

U = 1 4 _0 _ i < j q_i q_j r_ ij (NCERT Eq. 2.20)

• For N = 2 : 1 pair → U = kq_1q_2/r_ 12
• For N = 3 : 3 pairs → add up carefully
• For N = 4 identical charges q at corners of a square (side a ): U = k q^2 ( 4 a + 2 a 2 ) — 4 side pairs + 2 diagonal pairs
• The total PE can be positive, negative, or zero depending on signs and arrangement

PE of a Dipole in an External Field

A dipole in a uniform electric field has potential energy that depends on the angle between the dipole moment and the field. This is how molecules align in electric fields.

A dipole with moment p in a uniform electric field E experiences a torque = p E that tries to align it with the field. The potential energy of this configuration is:

U = - p E = -pE (NCERT Eq. 2.26)

• = 0° (aligned with E ): U = -pE — minimum PE, stable equilibrium
• = 90° (perpendicular): U = 0 — this is the reference orientation
• = 180° (anti-aligned): U = +pE — maximum PE, unstable equilibrium
• Torque: = pE — maximum when perpendicular, zero when aligned or anti-aligned
• The dipole naturally rotates to minimize PE → aligns with the field

Quick recall

Work done to rotate a dipole from = 0° to = 180° in a uniform field?

W = U = U_ 180° - U_ 0° = (+pE) - (-pE) = 2pE . This is the maximum possible work, corresponding to full reversal against the field.

Conductors & Electrostatics

Inside a conductor in electrostatic equilibrium, the electric field is zero, all excess charge sits on the surface, and the entire conductor is an equipotential body. These properties enable electrostatic shielding — Faraday cages protect from lightning.

E_ inside = 0, E_ surface = _0

Properties of Conductors

In electrostatic equilibrium, conductors have remarkable properties: zero internal field, surface-only charge, and uniform potential throughout.

When a conductor reaches electrostatic equilibrium (no current flowing), the free charges have arranged themselves to satisfy several remarkable conditions simultaneously. These are not independent rules — they all follow from one fact: charges are free to move.

• Rule 1: E = 0 inside the conductor — if it weren't, free charges would move until it becomes zero
• Rule 2: All excess charge resides on the outer surface only
• Rule 3: E at the surface is perpendicular to the surface: E = / _0
• Rule 4: The entire conductor (surface + interior) is an equipotential body
• Rule 5: In a cavity inside a conductor with no charge, E = 0 (electrostatic shielding)

E_ surface = _0 (NCERT Eq. 2.35)

Here is the surface charge density — charge per unit area at that point. On a conductor with irregular shape, is highest at sharp points (where curvature is greatest). This is why lightning rods have pointed tips.

Electrostatic Shielding

A conducting shell shields its interior from external electric fields. This is the Faraday cage effect — and it's why your car is safe in a lightning storm.

If a conductor has a cavity (hollow region inside) and there is NO charge inside the cavity, then the electric field in the cavity is exactly zero — regardless of what's happening outside. This is electrostatic shielding or the Faraday cage effect.

• External field induces charges on the outer surface of the conductor, but nothing penetrates into the cavity
• The cavity is at the same potential as the conductor — it's a field-free zone
• Shielding works even if the conductor is NOT grounded (just needs to be closed)
• If there IS a charge q inside the cavity, a charge -q appears on the inner surface and +q on the outer surface
• The reverse is NOT true: a charge inside a cavity DOES produce a field outside (the outer surface carries induced charge)

Quick recall

A hollow conducting sphere has a charge +Q on it. What's the field inside the cavity?

Zero. All +Q sits on the outer surface. The cavity has no charge on its inner surface, and by Gauss's law, the field inside a Gaussian surface enclosing no charge is zero.

Capacitors & Capacitance

A capacitor stores charge and energy using two conductors separated by an insulator. Capacitance C measures how much charge it can hold per volt. Every phone, laptop, and defibrillator relies on capacitors.

C = Q V = _0 A d

What is a Capacitor?

A capacitor is a device that stores electric charge and energy. It consists of two conductors separated by an insulator, and its capacity to hold charge is measured by capacitance C.

A capacitor consists of two conductors (called plates) separated by an insulating medium (dielectric or vacuum). When connected to a battery, one plate acquires charge +Q and the other -Q . The ratio of charge to potential difference defines capacitance.

C = Q V (NCERT Eq. 2.36)

• SI unit of capacitance → Farad (F) = 1 C/V
• 1 Farad is HUGE — common capacitors are in ( 10^ -6 ), nF ( 10^ -9 ), or pF ( 10^ -12 )
• Capacitance depends on geometry (shape, size, separation) and medium — NOT on Q or V
• The net charge on a capacitor is always zero ( +Q on one plate, -Q on the other)
• Named after Michael Faraday (1791–1867)

Parallel Plate Capacitor

Two large parallel conducting plates separated by distance d. The capacitance depends on plate area, separation, and the medium between them.

For a parallel plate capacitor with plate area A , separation d , and vacuum between the plates, the uniform electric field between the plates is E = / _0 = Q/( _0 A) . The potential difference is V = Ed .

C = _0 A d (NCERT Eq. 2.41)

• C A — larger plates → more capacitance
• C 1/d — smaller gap → more capacitance
• Field between plates is uniform: E = V/d
• Field outside the plates is approximately zero (edge effects ignored)
• This formula assumes A d^2 (plate dimensions much larger than separation)

Quick recall

A parallel plate capacitor has plates of area 100 cm^2 separated by 1 mm . Find C .

C = 8.85 10^ -12 100 10^ -4 1 10^ -3 = 8.85 10^ -11 F = 88.5 pF . Tiny — showing why 1 Farad is enormous.

Energy Stored in a Capacitor

A charged capacitor stores energy in the electric field between its plates. This energy can be released rapidly — that's how camera flashes and defibrillators work.

When a capacitor is charged, work is done by the battery against the electric field. This work is stored as electrostatic energy in the field between the plates. The three equivalent expressions are:

U = 1 2 CV^2 = 1 2 QV = Q^2 2C (NCERT Eq. 2.53)

• Energy is stored in the electric field, not on the plates
• Energy density (energy per unit volume) of the field: u = 1 2 _0 E^2
• Total energy = energy density × volume between plates: U = 1 2 _0 E^2 Ad
• When a capacitor discharges, energy is converted to heat, light, or work
• Battery delivers energy QV ; capacitor stores 1 2 QV ; the other half is lost as heat in the resistance

Combinations of Capacitors

Capacitors can be connected in series (same charge, voltages add) or parallel (same voltage, charges add). Knowing which quantity is shared is the key to solving any combination problem.

Series: 1 C_ eq = 1 C_i Parallel: C_ eq = C_i

Capacitors in Series

In series, each capacitor stores the same charge Q, but the voltage divides across them. The equivalent capacitance is LESS than the smallest individual capacitance.

When capacitors are connected end-to-end (series), the charge on each capacitor is the same (because the intermediate conductors are isolated). The voltages add up to the total applied voltage.

1 C_ eq = 1 C_1 + 1 C_2 + 1 C_3 + (NCERT Eq. 2.45)

• Same Q on all capacitors — this is the defining feature of series
• Voltages add: V = V_1 + V_2 + V_3 +
• C_ eq is always less than the smallest individual capacitance
• For two capacitors: C_ eq = C_1 C_2 C_1 + C_2 (product over sum)
• For N identical capacitors of capacitance C : C_ eq = C/N

Quick recall

Three capacitors of 2 , 3 , and 6 are in series. Find C_ eq .

1/C_ eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 . So C_ eq = 1 . Less than the smallest ( 2 ) — as expected for series.

Capacitors in Parallel

In parallel, each capacitor has the same voltage across it, but they can store different amounts of charge. The equivalent capacitance is the sum of all individual capacitances.

When capacitors are connected with both their plates joined together (parallel), the voltage across each is the same. Each stores charge independently, and the total charge is the sum.

C_ eq = C_1 + C_2 + C_3 + (NCERT Eq. 2.49)

• Same V across all capacitors — this is the defining feature of parallel
• Charges add: Q = Q_1 + Q_2 + Q_3 +
• C_ eq is always greater than the largest individual capacitance
• Charge distributes proportionally: Q_i = C_i V — bigger capacitor gets more charge
• For N identical capacitors of capacitance C : C_ eq = NC

Mixed Capacitor Circuits

Real circuits often have both series and parallel combinations. The strategy: identify which capacitors are in series or parallel, simplify step by step, then work backwards to find individual charges and voltages.

For mixed circuits, the approach is systematic: (1) Identify groups that are purely in series or purely in parallel, (2) Replace each group with its equivalent, (3) Repeat until you have a single equivalent capacitance, (4) Work backwards to find Q and V for each capacitor.

Solving Mixed Capacitor Circuits

1. Draw the circuit clearly and label all capacitors with their values
2. Identify the innermost series or parallel group that can be simplified
3. Replace it with its equivalent capacitance using the appropriate formula
4. Repeat until only one equivalent capacitor remains
5. Find total charge: Q_ total = C_ eq V_ total
6. Work backwards: series capacitors share Q; parallel capacitors share V

• Wheatstone bridge configuration: if the ratio condition C_1/C_2 = C_3/C_4 is met, the bridge capacitor carries no charge and can be removed
• Symmetry: in symmetric circuits, capacitors at equivalent positions have equal charges — use this to simplify
• Always check: total charge from battery = charge on the equivalent capacitance
• Energy conservation: total energy stored = sum of energies in individual capacitors

Dielectrics & Capacitors

Inserting a dielectric between capacitor plates increases capacitance by a factor K (dielectric constant). This happens because the dielectric reduces the net electric field. The Van de Graaff generator uses capacitor principles to build up millions of volts.

C = KC_0 = K _0 A d

Effect of Dielectric on Capacitance

A dielectric is an insulator that, when placed in an electric field, gets polarized and reduces the net field. This allows the capacitor to store MORE charge at the same voltage — increasing capacitance.

When a dielectric slab with dielectric constant K (also called or _r ) is inserted between the plates of a capacitor, the molecules of the dielectric get polarized. This creates a field opposing the original field, reducing the net field to E_0/K .

C = KC_0 = K _0 A d (NCERT Eq. 2.56)

• K 1 always ( K = 1 for vacuum, K 1 for air, K = 80 for water)
• With dielectric, C increases by factor K
• Net field inside dielectric: E = E_0/K (reduced)
• For a dielectric slab of thickness t < d : C = _0 A d - t + t/K
• A conductor ( K = ) inserted: C = _0 A/(d - t) — even larger increase

Battery Connected vs Disconnected

What happens when you change plate separation or insert a dielectric depends critically on whether the battery is still connected or has been removed. Master both cases.

Every NEET capacitor problem falls into one of two categories based on what stays constant during the change. This determines everything.

• With battery, inserting dielectric: energy INCREASES (battery does work to push extra charge)
• Without battery, inserting dielectric: energy DECREASES (the dielectric is pulled in, doing work)
• The dielectric is always attracted into the capacitor (force acts to increase C )
• If only part of the dielectric is inserted, the system behaves like two capacitors in parallel

Quick recall

A 10 capacitor is charged to 100 V and then disconnected. A dielectric with K = 5 is inserted. Find new V and U .

Q = CV = 10 10^ -6 100 = 10^ -3 C (conserved). New C = 5 10 = 50 . New V = Q/C' = 10^ -3 /50 10^ -6 = 20 V . New U = 1 2 Q^2/C' = 1 2 10^ -6 /50 10^ -6 = 0.01 J (was 0.05 J — energy decreased by factor 5).

Van de Graaff Generator

A Van de Graaff generator uses the principle that charge always moves to the outer surface of a conductor to continuously accumulate charge on a large hollow sphere, building up enormous potential.

The Van de Graaff generator exploits two key electrostatic principles: (1) charge placed inside a hollow conductor moves to the outer surface, and (2) this process can be repeated indefinitely because the interior surface always stays at the same potential as the sphere.

How the Van de Graaff Generator Works

1. A motorized insulating belt runs between two pulleys — one at the base, one inside the large metal sphere
2. At the base, a sharp comb (spray comb) at high voltage ionizes air and sprays positive charge onto the belt
3. The belt carries the charge upward into the interior of the hollow metal sphere
4. Inside the sphere, a second comb (collecting comb) picks up the charge from the belt
5. Since charge always moves to the OUTER surface of a conductor, the collected charge migrates to the outside of the sphere
6. The process repeats, accumulating more and more charge — building potential up to millions of volts

• Potential of the sphere: V = Q 4 _0 R — keeps increasing as Q increases
• Maximum voltage limited by dielectric breakdown of surrounding air ( 3 10^6 V/m )
• For a sphere of radius R : max V 3 10^6 R volts
• Large spheres in vacuum or high-pressure gas can reach ~20 million volts
• Used in nuclear physics to accelerate charged particles for collision experiments

Quick recall

Why does charge sprayed inside the sphere always move to the outside?

Because inside a conductor, E = 0 in equilibrium. Any charge placed inside creates a field that pushes the charge to the outer surface. Once on the outer surface, the interior returns to E = 0 . This is why you can keep adding charge — the inside is always 'ready' for more.