JEE Physics Electrostatic Potential and Capacitance Notes Set 02

Capacitance and Parallel Plate Capacitors

1. A capacitor stores charge \(Q\) when a potential difference \(V\) is applied: \(Q=CV\). Capacitance depends only on geometry and the medium, never on the charge or the metal used.
2. Parallel plates: \(C=\dfrac{K\varepsilon_0A}{d}\). Spherical conductor of radius \(a\): \(C=4\pi\varepsilon_0a\). Concentric spheres (inner \(a\), outer \(b\), outer earthed): \(C=4\pi\varepsilon_0\dfrac{ab}{b-a}\). Cylindrical capacitor: \(C=\dfrac{2\pi\varepsilon_0L}{\ln(b/a)}\), and the field falls as \(E\propto 1/r\).
3. Two plates with charges \(Q_1\) and \(Q_2\) placed face to face carry \(\dfrac{Q_1-Q_2}{2}\) on the facing surfaces (with opposite signs) and \(\dfrac{Q_1+Q_2}{2}\) on each outer surface.
4. The breakdown field of air is about \(3\times10^6\ \text{V/m}\); this limits the charge a conductor of given size can hold.

Energy and Force

1. Stored energy \(U=\tfrac12CV^2=\dfrac{Q^2}{2C}=\tfrac12QV\). Energy density \(u=\tfrac12\varepsilon_0E^2\) (multiply by \(K\) inside a dielectric).
2. Force between the plates \(F=\dfrac{Q^2}{2\varepsilon_0A}=\tfrac12\varepsilon_0AE^2\); the factor one half arises because each plate sits in the field of the other plate only.
3. Charging from a battery: the battery supplies \(QV\), half is stored and half is lost as heat, whatever the resistance.
4. Joining a charged and an uncharged capacitor: charge is conserved, common potential \(V=\dfrac{C_1V_1}{C_1+C_2}\), and the fraction of energy lost is \(\dfrac{C_2}{C_1+C_2}\). In general the loss when two charged capacitors are joined with like plates together is \(\dfrac{C_1C_2}{2(C_1+C_2)}(V_1-V_2)^2\).

Dielectrics and Slabs

A slab of thickness \(t\) and constant \(K\) in a gap \(d\) behaves like an air gap \(d-t+\dfrac{t}{K}\), so \(C=\dfrac{\varepsilon_0A}{d-t+t/K}\). If the slab fills only part of the area, treat the two regions as parallel capacitors; if layers lie one above the other, treat them as series capacitors.

Combinations

1. Series: \(\dfrac{1}{C}=\sum\dfrac{1}{C_i}\), same charge on each, voltage divides inversely with capacitance.
2. Parallel: \(C=\sum C_i\), same voltage on each, charge divides in proportion to capacitance.
3. Balanced bridge: if \(\dfrac{C_1}{C_2}=\dfrac{C_3}{C_4}\), the capacitor across the bridge carries no charge and can be removed.
4. For non-series-parallel networks give each internal node a potential, write that the net charge on every isolated node is conserved, and solve the linear equations. Alternate plates of a stack joined together form parallel capacitors, one for every gap.
5. Capacitor strings made of units with a rated voltage: choose the number in series from the voltage and the number of strings from the capacitance.

Charging and Discharging Through a Resistor

1. Charging: \(q=q_0(1-e^{-t/RC})\), \(I=\dfrac{V}{R}e^{-t/RC}\). Discharging: \(q=q_0e^{-t/RC}\). The time constant is \(\tau=RC\).
2. At \(t=0\) an uncharged capacitor acts like a wire, and after a long time a capacitor acts like an open circuit, so in the steady state no current flows in a branch containing a capacitor.
3. A plot of \(\log I\) against \(t\) for discharge is a straight line of slope \(-\dfrac{1}{RC}\); the intercept gives \(\log\dfrac{V}{R}\).
4. Time to reach 99% of the final charge is \(RC\ln100\approx4.6RC\); the charge halves in \(0.693RC\) during discharge.

Worked Examples

1. Two capacitors \(3\ \mu\text{F}\) and \(6\ \mu\text{F}\) in series across \(9\ \text{V}\): \(C=2\ \mu\text{F}\), \(Q=18\ \mu\text{C}\) on each, so the voltages are \(6\ \text{V}\) and \(3\ \text{V}\).
2. A \(10\ \mu\text{F}\) capacitor at \(100\ \text{V}\) is joined to an uncharged \(15\ \mu\text{F}\) capacitor. Charge \(=1000\ \mu\text{C}\) is shared by \(25\ \mu\text{F}\), so \(V=40\ \text{V}\); energy falls from \(50\ \text{mJ}\) to \(20\ \text{mJ}\), a loss of 60%, which equals \(\dfrac{15}{25}\).
3. A \(2\ \text{mm}\) thick slab with \(K=5\) is placed in a \(4\ \text{mm}\) gap: effective gap \(=4-2+0.4=2.4\ \text{mm}\), so the capacitance rises by the factor \(\dfrac{4}{2.4}=\dfrac{5}{3}\).