Electrostatic Potential and Capacitance

Electrostatic potential energy

• Coulomb force is conservative: work done in moving a charge between two points is independent of the path and zero over a closed loop.
• Potential energy difference $\Delta U=U_P-U_R=W_{RP}$, the work done by an external force in moving $q$ from $R$ to $P$ without acceleration.
• $U$ is fixed only up to an additive constant. Standard choice: $U=0$ at infinity, so $U_P=W_{\infty P}$.

Electrostatic potential

• $V$ at a point = work done per unit positive charge by an external agent in bringing it from infinity to that point, without acceleration.
• $V_P-V_R=\dfrac{U_P-U_R}{q}$. Potential is a scalar; add contributions algebraically, with sign.
• SI unit volt: $1\ \text{V}=1\ \text{J C}^{-1}=1\ \text{N C}^{-1}\text{m}$. A quantity whose unit is $\text{N C}^{-1}\text{m}$ is electrostatic potential.
• Work by external agent from $A$ to $B$ is $q(V_B-V_A)$; work by the field is $q(V_A-V_B)$. Around any closed loop in an electrostatic field, $W=0$.
• $1\ \text{eV}$ is the energy gained by an electron accelerated through $1\ \text{V}$: $1.6\times10^{-19}\ \text{J}$.

Potential due to a point charge

• $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$, positive for $Q>0$, negative for $Q<0$, zero at infinity.
• $V\propto\dfrac1r$ (hyperbola), $E\propto\dfrac{1}{r^{2}}$ — the potential curve falls off more slowly than the field curve. On a $V$ versus $1/r$ plot the graph is a straight line through the origin of slope $\dfrac{Q}{4\pi\varepsilon_0}$, so slope ratio gives charge ratio.
• Potential due to a charge at its own location is undefined (infinite).

Potential due to an electric dipole

• Dipole potential depends on $r$ and on $\theta$; it is axially symmetric about $\vec p$.
• It falls as $1/r^{2}$, faster than the $1/r$ of a single charge, because the total charge is zero.

Potential due to a system of charges

• Superposition: $V=\dfrac{1}{4\pi\varepsilon_0}\displaystyle\sum_i\dfrac{q_i}{r_{iP}}$ — an algebraic sum, never a vector sum.
• Charged conducting sphere or thin shell of radius $R$, charge $q$: $V=\dfrac{kq}{r}$ outside, and $V=\dfrac{kq}{R}$ everywhere inside and on the surface. $E$ is discontinuous across the surface; $V$ is continuous.
• $N$ droplets of radius $r$ at potential $V$ each, coalescing: $R=N^{1/3}r$, $V'=N^{2/3}V$.
• Two spheres joined by a wire come to a common potential, not a common charge: $\dfrac{q_1'}{q_2'}=\dfrac{r_1^{2}}{r_2^{2}}$ if the shared surface density is uniform, i.e. $\sigma\propto\dfrac1r$ for equal potentials.
• Zero-potential point for $q$ and $-2q$ separated by $d$: at $d/3$ from $q$ (between them) — and one more point outside, on the $q$ side.

Equipotential surfaces

• Definition: a surface on which the potential has the same value at every point.
• Properties CBSE asks to "write two of": no work is done in moving a charge over the surface; $\vec E$ is everywhere normal to the surface; two equipotential surfaces never intersect; closer spacing means a stronger field.
• Why normal: a tangential component of $\vec E$ would do work along the surface, but $W=q(V_B-V_A)=0$ there. So the tangential component must vanish.

• Surfaces of a point charge are not equidistant: $V\propto1/r$, so for equal steps in $V$ the spacing grows with $r$.
• Uniform field with magnitude increasing along $-z$: still parallel planes normal to $z$, but crowded together towards $-z$.

Relation between field and potential

• Bringing a unit positive charge through $\delta l$ against $\vec E$ from surface $B$ ($V$) to surface $A$ ($V+\delta V$) costs work $E\,\delta l$, so $E\,\delta l=-\delta V$.
• $E=-\dfrac{dV}{dr}$: the field is the negative gradient of the potential.
• Conclusion (i): $\vec E$ points in the direction of steepest decrease of $V$. Conclusion (ii): its magnitude is the rate of change of $V$ per unit displacement normal to the equipotential surface.
• Reverse direction: $V_B-V_A=-\displaystyle\int_A^B\vec E\cdot d\vec r$.

Potential energy of a system of charges

• $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ for a pair; sum over all $\dfrac{n(n-1)}{2}$ pairs for $n$ charges.
• $U>0$ for like charges (work must be done to push them together), $U<0$ for unlike charges.
• $U$ depends only on the final configuration, not on the order of assembly.
• Work needed to separate the charges to infinity $=-U$.
• Standard results: equilateral triangle of side $a$ with $q,2q,-3q$ gives $U=\dfrac{-7q^{2}}{4\pi\varepsilon_0 a}$; with $Q,q,-q$ on a triangle of side $L$, $U=-\dfrac{kq^{2}}{L}$ (the $Q$ terms cancel).

Potential energy in an external field

• Single charge: $U=qV(\vec r)$, with $V$ the external potential only.
• Two charges in an external field: $U=q_1V(\vec r_1)+q_2V(\vec r_2)+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$.
• Dipole in a uniform field: net force zero, torque $\vec\tau=\vec p\times\vec E$, $U=-\vec p\cdot\vec E=-pE\cos\theta$.

• Work to rotate from $\theta_1$ to $\theta_2$: $W=pE(\cos\theta_1-\cos\theta_2)$. Rotating from parallel through $90^\circ$ costs $pE$; from $90^\circ$ to $180^\circ$ also costs $pE$.

Electrostatics of conductors — the six results

• Inside a conductor the electrostatic field is zero. Free charges rearrange until it is.
• At the surface of a charged conductor $\vec E$ is normal to the surface at every point; a tangential component would move surface charges.
• The interior can carry no excess charge in the static state; all excess charge sits on the outer surface (Gauss's law on a vanishingly small interior surface).
• $V$ is constant throughout the volume and on the surface — the whole conductor is one equipotential.
• Just outside the surface, $\vec E=\dfrac{\sigma}{\varepsilon_0}\hat n$ (pill-box Gaussian surface).
• Electrostatic shielding: the field in a charge-free cavity is zero, whatever lies outside. Shielding works one way only — charges placed inside the cavity are not screened from the outside.

Dielectrics and polarisation

• Non-polar molecules ($\text{O}_2$, $\text{H}_2$): centres of positive and negative charge coincide, no permanent dipole moment; a field induces one.
• Polar molecules ($\text{HCl}$, $\text{H}_2\text{O}$): permanent dipole moment, randomly oriented by thermal agitation; a field aligns them.
• Polarisation $\vec P$ = dipole moment per unit volume; for linear isotropic dielectrics $\vec P=\varepsilon_0\chi_e\vec E$.
• A uniformly polarised slab is equivalent to two sheets of bound charge $\pm\sigma_P$ on the faces normal to $\vec E$; there is no net volume charge.
• The bound-charge field opposes the applied field, so $E=\dfrac{E_0}{K}$ with $K>1$.

Capacitors and capacitance

• A capacitor is two conductors separated by an insulator, carrying $+Q$ and $-Q$; its total charge is zero and $Q$ means the charge on one plate.
• $C=\dfrac{Q}{V}$; $C$ is fixed by shape, size, separation and the medium, and does not change with $Q$ or $V$.
• Slope of a $q$–$V$ graph is $C$: the steeper line is the larger capacitance.
• A large $C$ stores a large $Q$ at a modest $V$, keeping the field below the dielectric strength of the medium ($3\times10^{6}\ \text{V m}^{-1}$ for air) so the charge does not leak.

The parallel plate capacitor

• Field of each sheet: $\dfrac{\sigma}{2\varepsilon_0}$. Between the plates they add to $\dfrac{\sigma}{\varepsilon_0}$; outside they cancel to zero.
• $V=Ed=\dfrac{Qd}{\varepsilon_0A}$ and $C=\dfrac{\varepsilon_0A}{d}$, a purely geometric result valid for $d^{2}\ll A$ (fringing ignored).
• $1\ \text{F}$ is enormous: with $d=1\ \text{cm}$ it needs plates about $30\ \text{km}$ on a side.
• Removing one plate halves the field, hence halves the force on a charge between the plates.
• Charging across a dc battery: a momentary current piles electrons on one plate and leaves an equal positive charge on the other; the growing potential difference opposes further charging, and the current stops when the plate voltage equals the battery voltage.

Effect of a dielectric

• Full slab: $C=\dfrac{K\varepsilon_0A}{d}=KC_0$. Partial slab of thickness $t$: $C=\dfrac{\varepsilon_0A}{d-t+t/K}$; a conducting slab is the $K\to\infty$ case, $C=\dfrac{\varepsilon_0A}{d-t}$.
• Slabs stacked across the gap are in series; slabs sitting side by side across the area are in parallel.

Combination of capacitors

• Series: charge the same on each, $V=V_1+V_2+\cdots$, $\dfrac{1}{C_s}=\displaystyle\sum\dfrac{1}{C_i}$; the equivalent is smaller than the smallest member.
• Parallel: potential the same across each, $Q=Q_1+Q_2+\cdots$, $C_p=\displaystyle\sum C_i$; the equivalent is larger than the largest member.
• Three equal capacitors $C$: all parallel gives $3C$, all series gives $C/3$, two in series with the third in parallel gives $\dfrac{3C}{2}$.
• Balanced bridge: if the bridge condition holds, the bridging capacitor carries no charge and can be removed before reducing the network.

Energy stored in a capacitor

• Building the charge bit by bit: $W=\displaystyle\int_0^Q\dfrac{Q'}{C}dQ'=\dfrac{Q^{2}}{2C}$, hence $U=\dfrac12CV^{2}=\dfrac12QV=\dfrac{Q^{2}}{2C}$.
• Energy density $u=\dfrac12\varepsilon_0E^{2}$ — a general result for any field configuration, not only for parallel plates.
• In series the same $Q$ flows, so $U=\dfrac{Q^{2}}{2C_{\text{eff}}}=\dfrac{Q^{2}}{2C_1}+\dfrac{Q^{2}}{2C_2}+\cdots$: total energy equals the sum of individual energies (the same holds in parallel with $\dfrac12CV^{2}$).
• Sharing charge between an identical charged and uncharged capacitor halves the stored energy; the missing half is dissipated as heat and electromagnetic radiation by the transient current. Charge is conserved, energy is not.

Traps that cost marks

• Potential is a scalar: add with signs, never with the parallelogram law. Field is a vector: add with components.
• $E=0$ does not mean $V=0$. Inside a charged shell $E=0$ but $V=\dfrac{kq}{R}\neq0$. Conversely $V=0$ does not mean $E=0$ — on the equatorial plane of a dipole $V=0$ but $E\neq0$.
• In $U=qV(\vec r)$, $V$ must exclude the potential of $q$ itself, which is infinite at its own location.
• Do not quote $V=Ed$ for a non-uniform field; integrate $-\int\vec E\cdot d\vec r$ instead.
• Write $U=-pE\cos\theta$, not $+pE\cos\theta$; state that $U=0$ is chosen at $\theta=90^\circ$ or you lose the justification mark.
• Decide first whether the battery is still connected. Connected means $V$ is fixed; disconnected means $Q$ is fixed. Every "how is it affected" question turns on this.
• With the battery disconnected, $E=\dfrac{Q}{\varepsilon_0A}$ does not depend on the plate separation — pulling the plates apart changes $C$, $V$ and $U$, but not $E$.
• In series the charge is the same, not the voltage; in parallel the voltage is the same, not the charge. Reversing the two is the commonest slip in capacitor numericals.
• The dielectric constant multiplies capacitance, so $K$ goes in the numerator of $C=\dfrac{K\varepsilon_0A}{d}$, but it divides the field, $E=\dfrac{E_0}{K}$.
• Two equipotential surfaces can never cross — a point would then have two potentials.
• A force question and a potential-energy question about the same array are different: force needs vector addition at $120^\circ$ or along a diagonal, energy needs the scalar sum over pairs.
• Both a proton and an electron in the same field feel the same magnitude of force ($F=qE$ with equal $\lvert q\rvert$); their potential energies $U=qV$ differ in sign.
• Convert $\mu\text{F}$, $\text{pF}$, cm and mm to SI before substituting; most lost numerical marks are unit slips, not physics.