Electrostatic Potential and Capacitance

The chapter has two connected halves. The first half (potential and potential energy) builds the scalar description of the electric field: potential energy and potential from work done, potential of a point charge and a dipole, potential of a system of charges and of a charged spherical shell, equipotential surfaces and the field-potential relation, potential energy of a system of charges, potential energy in an external field including a dipole in a uniform field, and the electrostatics of conductors (zero interior field, surface charge, constant potential, electrostatic shielding). The second half (capacitance) turns this into a device: dielectrics and polarisation, capacitors and capacitance, the parallel plate capacitor, the effect of a dielectric on capacitance, series and parallel combinations, and the energy stored in a capacitor together with the idea of energy density in the electric field. Worked examples throughout reuse NCERT's own numbering (Examples 2.1 to 2.10) so you can cross-check against your textbook.

From work done to potential energy

Coulomb force, like gravity, is a conservative force — the work done moving a charge between two points does not depend on the path taken, only on the endpoints. Bringing a positive test charge $q$ from a point $R$ to a point $P$ against the repulsive field of a source charge $Q$ requires an external force that does work $W_{RP}$, and this work is stored entirely as potential energy, since removing the external force lets the field convert that stored energy back into kinetic energy.

where: $U_P, U_R$ are the potential energies of the test charge at $P$ and $R$, and $W_{RP}$ is the work done by an external agent (equal and opposite to the electric force) in moving the charge from $R$ to $P$ without giving it any net kinetic energy. Because only differences in $U$ ever enter physics, potential energy is defined only up to an arbitrary additive constant — the natural choice is to set $U=0$ at infinity, which then fixes $U$ at any point $P$ as the work done bringing the charge from infinity to $P$.

Electrostatic potential

Dividing potential energy by the test charge $q$ removes the dependence on how big a test charge you happened to use, leaving a property of the field alone: the electrostatic potential $V$.

where: $V_P, V_R$ are the potentials at $P$ and $R$ in volts ($1\ \text{V} = 1\ \text{J/C}$). With $V=0$ chosen at infinity, $V$ at any point is the work done by an external agent in bringing a unit positive charge from infinity to that point, without acceleration. Potential, like potential energy, is a scalar — no direction to track, only a sign and a magnitude — which is exactly what makes it so much easier to add up over many charges than the vector field $\vec E$.

Potential due to a point charge

Placing charge $Q$ at the origin and integrating the Coulomb force along a radial path from infinity to a point at distance $r$ gives the potential everywhere around a point charge, and this is the seed formula for the whole chapter.

where: $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space, and $r$ is the distance from the charge (full derivation below). Equation (3) holds for either sign of $Q$: positive charge gives positive potential everywhere, negative charge gives negative potential everywhere, consistent with $V=0$ at infinity. Unlike the field, which falls off as $1/r^2$, potential falls off more gently, as $1/r$.

Potential due to an electric dipole

An electric dipole is a charge $+q$ and $-q$ separated by a small distance $2a$, with dipole moment $\vec p$ of magnitude $q(2a)$ pointing from $-q$ to $+q$. Superposing the potentials of the two point charges and expanding for a point far from the dipole ($r \gg a$) gives a result strikingly different from a single charge.

where: $\theta$ is the angle between the position vector $\vec r$ of the field point and the dipole moment $\vec p$. The dipole's potential depends not just on distance but on direction — maximum and positive on the axis toward $+q$ ($\theta=0$), equally maximum but negative toward $-q$ ($\theta=\pi$), and exactly zero on the equatorial plane ($\theta=\pi/2$), since the two charges are equidistant there and cancel. It also falls off as $1/r^2$, one power of $r$ faster than a single charge's $1/r$, because the near-cancellation of $+q$ and $-q$ weakens it faster with distance.

Potential due to a system of charges, and a charged spherical shell

Because potential is a scalar built from a linear operation (division by $q$ of a linear work integral), it inherits the superposition principle exactly as the field does, but without any vector addition to worry about.

where: $r_{iP}$ is the distance from the $i$-th charge to the field point $P$. This scalar sum is why potential is often easier to compute than the field for a complicated arrangement — no components to resolve. Applying it to a uniformly charged spherical shell of charge $q$, radius $R$ (whose field is zero inside and behaves as a point charge outside) gives $V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ for $r\ge R$, and, since no work is needed through a zero field, the constant value $V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R}$ for every point inside, including the centre.

Equipotential surfaces and the field-potential relation

A surface on which $V$ has the same value everywhere is an equipotential surface; for a single point charge these are concentric spheres centred on the charge, and for a uniform field they are planes perpendicular to the field direction. The defining geometric fact links the scalar and vector pictures: the field is always perpendicular to the equipotential surface through any point, because any component of $E$ along the surface would mean work is needed to move a charge along a supposedly constant-potential surface — a contradiction.

where: $dV$ is the change in potential over an infinitesimal displacement $dl$ measured normal to the equipotential surfaces. The field points in the direction potential decreases most steeply (hence the minus sign — high to low potential), and its magnitude equals the rate of that decrease per unit distance. Closely packed equipotential surfaces mean a strong field; widely spaced ones mean a weak field — like closely spaced contour lines meaning steep terrain on a map.

Potential energy of a system of charges

Building up a configuration of point charges from infinity, one at a time, costs no work for the first charge (nothing else is there yet), but every subsequent charge must be moved through the potential created by all the charges already in place. Summing this over all pairs gives the total electrostatic potential energy of the configuration, independent of the order of assembly because the Coulomb force is conservative.

where: the sum runs over every distinct pair of charges, and $r_{ij}$ is the distance between charges $i$ and $j$. Same-sign charges give positive energy (work was needed against repulsion to bring them together); opposite signs give negative energy (the attractive force did positive work, so an external agent had to hold them back with negative work). The same expression applies term by term for any number of charges — square corners, hexagon vertices, whatever arrangement a problem gives you.

Potential energy in an external field, and the dipole case

A related but distinct question: given an external field $E$ (or potential $V$) produced by sources we are not tracking, what is the potential energy of a charge $q$ placed inside it? By the very definition of potential, the answer is immediate, and extending it to a dipole reveals the origin of the torque you met in Chapter 1.

where: the first form gives the energy of a charge $q$ at $\vec r$ in an external potential $V(\vec r)$; the second gives a dipole of moment $p$ at angle $\theta$ to a uniform field $E$, with zero energy chosen at $\theta=\pi/2$. Energy is lowest ($-pE$) when aligned with the field, highest ($+pE$) when anti-aligned, and the torque $\vec\tau=\vec p\times\vec E$ from Chapter 1 is exactly what rotates the dipole downhill toward this minimum — aligned is stable, anti-aligned is not.

Electrostatics of conductors

A conductor's free electrons redistribute themselves until electrostatic equilibrium is reached, enforcing a short, memorable list of results examiners test relentlessly. Inside a conductor the field is always zero (otherwise free charges would still be drifting); excess charge resides only on the outer surface (Gauss's law on any interior volume gives zero enclosed charge, since the field on its boundary is zero); the whole conductor sits at one constant potential (zero field means zero work to move a test charge inside or on it); and just outside a charged surface the field is normal to it, fixed by the local surface charge density.

where: $\sigma$ is the surface charge density at that point and $\hat n$ is the outward unit normal. A further consequence, electrostatic shielding, is that a charge-free cavity inside any conductor has exactly zero field inside it, whatever charge sits on the conductor or whatever external field surrounds it — the reason sensitive electronics are built inside metal enclosures (Faraday cages).

Dielectrics, polarisation, and the capacitor

Dielectrics are insulators with no free charge carriers; placed in an external field, they cannot cancel the field the way a conductor does, but they weaken it. Non-polar molecules (like O$_2$) develop an induced dipole moment as the field stretches apart their charge centres; polar molecules (like H$_2$O), already carrying a permanent dipole moment, instead align with the field against thermal disruption. Either way the dielectric develops a net dipole moment per unit volume, the polarisation $P=\varepsilon_0\chi_e E$, equivalent to induced bound surface charges $\pm\sigma_P$ on its two faces, producing a field opposing (but not cancelling) the external one. This weakened-but-not-cancelled internal field is exactly what makes a capacitor — two conductors carrying charges $+Q$ and $-Q$, separated by an insulator — a practical way to store charge safely at a controlled, low potential difference.

where: $Q$ is the magnitude of charge on either conductor, $V$ is the potential difference between them, and $C$, the capacitance, depends purely on geometry (shape, size, separation) and the dielectric between them — never on $Q$ or $V$ themselves, since doubling $Q$ doubles both the field and $V$ in proportion. The SI unit is the farad, $1\ \text{F}=1\ \text{C}\,\text{V}^{-1}$; a farad is an enormous unit (a 1 F capacitor with a 1 cm gap would need plates roughly 30 km on a side), so practical capacitances are quoted in microfarads, nanofarads or picofarads ($\text{pF}=10^{-12}\,\text{F}$).

The parallel plate capacitor and the effect of a dielectric

For two large parallel plates of area $A$ separated by a small distance $d$ (so $d^2 \ll A$ and fringing at the edges can be ignored), the fields of the two oppositely charged sheets cancel outside the plates and add uniformly between them, giving a capacitance that depends only on geometry and on whatever fills the gap.

where: $K$ is the dielectric constant of whatever fills the gap ($K=1$ for vacuum, $K>1$ for every real dielectric). The dielectric's induced bound charge $\sigma_P$ partially cancels the free charge's field, so for the same free charge $Q$, the field and hence $V$ inside drop by the factor $K$, and since $C=Q/V$, a smaller $V$ for the same $Q$ means a larger $C$. This is why real capacitors sandwich a dielectric between their plates — it multiplies the capacitance for the same size, and raises the breakdown voltage too.

Combining capacitors: series and parallel

Real circuits rarely use one capacitor alone, so the chapter gives two combination rules, mirror images of the series and parallel rules you will meet for resistors in the very next chapter (though with the two roles of $Q$ and $V$ swapped).

where: in a series combination, charge $Q$ has nowhere else to go, so every capacitor carries the same charge $Q$ while potential splits across them, $V=V_1+V_2+\cdots$, producing the reciprocal-sum rule; in a parallel combination, every capacitor shares the same $V$ while charge splits, $Q=Q_1+Q_2+\cdots$, giving the direct-sum rule. Quick sanity check: series always gives a smaller net capacitance than the smallest individual one; parallel always gives a larger one than the largest.

Energy stored in a capacitor

Charging a capacitor means moving charge, bit by bit, from one plate to the other against a growing potential difference; integrating the work done at each infinitesimal step gives the total energy stored, which can be written in three equivalent forms depending on which two of $Q$, $V$, $C$ a problem gives you.

where: $Q$ is the final charge and $V$ the final potential difference. This energy is not lost when the capacitor is disconnected — it stays stored, ready to be released, which is why a capacitor (unlike a resistor) can power a sudden discharge, as in a camera flash. Rewriting the parallel-plate case using the field $E=\sigma/\varepsilon_0$ shows the energy can equally be thought of as stored in the field itself, occupying volume $Ad$ at a density $u=\tfrac12\varepsilon_0 E^2$ — derived here for a parallel plate capacitor, but true in general for any electric field.

Board level - potential and work done bringing a charge from infinity

Calculate the potential at a point $P$ due to a charge of $4\times10^{-7}\ \text{C}$ located $9\ \text{cm}$ away. Hence find the work done in bringing a charge of $2\times10^{-9}\ \text{C}$ from infinity to $P$.

Step 1. Apply the point-charge formula, Equation (3), with $Q = 4\times10^{-7}\ \text{C}$ and $r = 9\times10^{-2}\ \text{m}$: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} = (9\times10^{9})\dfrac{4\times10^{-7}}{9\times10^{-2}} = 4\times10^{4}\ \text{V}$.

Step 2. Work done bringing charge $q = 2\times10^{-9}\ \text{C}$ from infinity to $P$ uses $W = qV$ directly, since $V$ is by definition the work per unit charge: $W = (2\times10^{-9})(4\times10^4) = 8\times10^{-5}\ \text{J}$.

Step 3. This work is path-independent, since electrostatic force is conservative: any infinitesimal path splits into a component along $r$ (which does work) and a component perpendicular to $r$ (which does zero work, since the force is radial). Answer: $V = 4\times10^4\ \text{V}$ and $W = 8\times10^{-5}\ \text{J}$, independent of the path taken.

Board level - equivalent capacitance and charge in a mixed network

A network of four $10\ \mu\text{F}$ capacitors is connected to a $500\ \text{V}$ supply: three of them, $C_1, C_2, C_3$, are in series with each other, and this series group is in parallel with the fourth, $C_4$. Find the equivalent capacitance and the charge on each capacitor.

Step 1. Combine the series group first, using Equation (12): $\dfrac{1}{C'} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3} = \dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10} = \dfrac{3}{10}\ \mu\text{F}^{-1}$, so $C' = \dfrac{10}{3}\ \mu\text{F} \approx 3.33\ \mu\text{F}$.

Step 2. This series group $C'$ sits in parallel with $C_4 = 10\ \mu\text{F}$, so the total is a direct sum: $C = C' + C_4 = 3.33 + 10 = 13.3\ \mu\text{F}$.

Step 3. In the series branch, the same charge $Q$ flows through $C_1$, $C_2$, $C_3$; the potential drop across the branch is the full $500\ \text{V}$, so $Q = C'\times500 = 1.7\times10^{-3}\ \text{C}$ on each. The parallel branch $C_4$ sees $500\ \text{V}$ directly, so $Q_4 = C_4\times500 = 5.0\times10^{-3}\ \text{C}$. Answer: equivalent capacitance $\mathbf{13.3\ \mu F}$; charge on each of $C_1, C_2, C_3$ is $\mathbf{1.7\times10^{-3}\ C}$, and on $C_4$ is $\mathbf{5.0\times10^{-3}\ C}$.

Competitive level (JEE / NEET) - energy "lost" when a charged capacitor shares charge with an uncharged one

A $600\ \text{pF}$ capacitor is charged by a $200\ \text{V}$ supply, then disconnected from the supply and connected to another uncharged $600\ \text{pF}$ capacitor. How much electrostatic energy is lost in the process, and where does it go?

Step 1. Initial charge and energy on the first capacitor before connection: $Q = CV = (600\times10^{-12})(200) = 1.2\times10^{-7}\ \text{C}$, and initial energy $U_i = \tfrac12 CV^2 = \tfrac12(600\times10^{-12})(200)^2 = 1.2\times10^{-5}\ \text{J}$.

Step 2. When connected, charge $Q$ redistributes between the two equal capacitors until both reach the same common potential $V'$; by charge conservation the total charge $Q$ splits equally, so each capacitor ends with $Q/2$, and since both have the same capacitance $C$, the common potential is $V' = \dfrac{Q/2}{C} = \dfrac{V}{2} = 100\ \text{V}$.

Step 3. Final energy is the sum over both capacitors, now each at $C=600\ \text{pF}$ and $V'=100\ \text{V}$: $U_f = 2\times\tfrac12 C V'^2 = C V'^2 = (600\times10^{-12})(100)^2 = 6\times10^{-6}\ \text{J}$. Alternatively, using the equivalent parallel capacitance $2C$ at $V'$: $U_f = \tfrac12(2C)(V')^2$, the same result.

Step 4. Energy lost is $\Delta U = U_i - U_f = 1.2\times10^{-5} - 6\times10^{-6} = 6\times10^{-6}\ \text{J}$, exactly half the initial energy. No charge is lost, but energy is lost because a transient current flows briefly while the potentials equalise, dissipating energy as heat in the wires (and briefly as radiation). Answer: energy lost is $\mathbf{6\times10^{-6}}$ J, exactly half the original $1.2\times10^{-5}$ J, dissipated as heat during the brief redistribution current. Timing note: under two minutes if you remember the shortcut. For two equal capacitors sharing charge, the common potential is always the average of what each would have alone (here, average of $200\ \text{V}$ and $0\ \text{V}$ is $100\ \text{V}$), and the general result for any two capacitors $C_1$ (charged to $V_1$) connected to $C_2$ (charged to $V_2$) is $\Delta U = \dfrac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2$ — memorising this formula turns a multi-step problem into a single substitution, a favourite JEE shortcut.

Competitive level (JEE / NEET) - capacitance with a partial dielectric slab

A parallel plate capacitor with plate separation $d$ has a dielectric slab of dielectric constant $K$ inserted between the plates, but the slab's thickness is only $\tfrac34 d$ (it does not fill the whole gap). Find the new capacitance in terms of the vacuum capacitance $C_0$.

Step 1. Split the gap into two regions in series: a dielectric-filled region of thickness $t=\tfrac34 d$ where the field is reduced to $E = E_0/K$ (with $E_0 = V_0/d$ the vacuum-case field), and a vacuum region of thickness $d-t = \tfrac14 d$ where the field remains the full $E_0$.

Step 2. The total potential difference is the field times distance, summed over the two regions: $V = E_0\left(\dfrac14 d\right) + \dfrac{E_0}{K}\left(\dfrac34 d\right) = E_0 d\left(\dfrac14 + \dfrac{3}{4K}\right) = E_0 d\cdot\dfrac{K+3}{4K}$. Recognising $E_0 d = V_0$, this is $V = V_0\cdot\dfrac{K+3}{4K}$.

Step 3. The free charge $Q_0$ on the plates is unchanged by inserting the slab (nothing was connected or disconnected during insertion in this idealisation), so the new capacitance is $C = \dfrac{Q_0}{V} = \dfrac{Q_0}{V_0\cdot\frac{K+3}{4K}} = \dfrac{4K}{K+3}\cdot\dfrac{Q_0}{V_0} = \dfrac{4K}{K+3}\,C_0$.

Step 4. Sanity-check the limits: if $K \to 1$ (slab is effectively vacuum), $C \to \dfrac{4}{4}C_0 = C_0$ as it must; if $t\to d$ (slab fills the whole gap, the special case in the chapter's own derivation of $C=KC_0$), a similar calculation with $t=d$ collapses cleanly to $C=KC_0$, confirming the partial-slab method reduces correctly to the full-slab formula. Answer: $\mathbf{C = \dfrac{4K}{K+3}\,C_0}$. Timing note: about ninety seconds once you recognise the pattern. The general method — split the gap into series regions of uniform field, sum $V = \sum E_i d_i$, then use $C=Q_0/V$ — handles any combination of slabs and air gaps a JEE question throws at you, including multiple slabs of different $K$ stacked in the same gap; always check the $K\to1$ and full-thickness limits as a thirty-second safety net before committing to an answer.

Confusing electric field and electric potential is the single most common source of errors — a point can have zero field but nonzero potential (the centre of two equal like charges) or zero potential but nonzero field (between a $+3q$ and a $-2q$ charge where potentials cancel but the field does not, Example 2.2). Always ask which quantity a question wants. Writing $V = Ed$ as if it applies everywhere is another frequent slip — it holds only for a uniform field between parallel plates; for a point charge or any non-uniform field, go back to $E=-dV/dl$ or integrate. Forgetting the factor of $\tfrac12$ in energy formulas is common under time pressure — students write $U=QV$ instead of $U=\tfrac12 QV$, forgetting that charge was moved gradually against a growing potential, not all at once against the final value. Assuming both charge and energy are conserved when two capacitors are connected is a related mistake — charge is always conserved, but energy is generally not (Worked example 3); do not assume $U_i=U_f$ just because nothing was "lost" in the everyday sense. Mixing up series and parallel rules for capacitors with the resistor rules of the next chapter trips up many students, since the roles are swapped: capacitors in series share the same charge, capacitors in parallel share the same voltage. Believing a dielectric cancels the field, rather than merely weakening it by the factor $K$, is a conceptual error worth flagging — a dielectric is not a conductor. Assuming a charged conductor's potential equals its field is a subtle Assertion-Reasoning trap — the field can be strong right at a highly charged surface while the potential is a modest, finite number; the two are related by geometry, not interchangeable.

• Define electrostatic potential energy and electrostatic potential from first principles, and explain why both are defined only up to an arbitrary additive constant.
• Write down and apply the potential of a point charge, a dipole, and a system of point charges, including the special case of a uniformly charged spherical shell inside and outside.
• Sketch equipotential surfaces for a point charge, a uniform field, and a dipole, and correctly state and use the relation $E=-dV/dl$.
• Compute the potential energy of a system of point charges, and separately the potential energy of a charge or a dipole placed in an external field.
• State all five key electrostatic properties of a conductor in equilibrium, and explain electrostatic shielding.
• Explain, at the molecular level, how polar and non-polar dielectrics polarise in an external field and why this weakens (without cancelling) the internal field.
• Derive the capacitance of a parallel plate capacitor, both with vacuum and with a dielectric filling all or part of the gap.
• Reduce any series-parallel network of capacitors to a single equivalent capacitance, and find the charge and voltage on each individual capacitor within it.
• Compute the energy stored in a charged capacitor in all three equivalent forms, and explain the energy density interpretation of the electric field.
• Correctly diagnose, in a mixed problem, whether a question is really asking about field, potential, potential energy, or capacitance — and avoid substituting one for another.

Ask the tutor: Want to work through a mixed series-parallel capacitor network with a dielectric slab in one branch, step by step, using your own numbers?