define capacitance, as applied to both isolated spherical conductors and to parallel plate capacitors

Capacitors and Capacitance (Cambridge AS & A‑Level Physics 9702 – Topic 19)

Learning Objectives

• Define capacitance, state its SI unit and the fundamental relation C = Q/V.
• Derive the capacitance of an isolated spherical conductor (vacuum and dielectric cases).
• Derive the capacitance of a parallel‑plate capacitor, including the effect of a dielectric (κ).
• Combine capacitors in series, in parallel and in mixed networks; calculate the equivalent capacitance.
• Write the three equivalent energy‑storage formulas and decide which is most convenient in a given situation.
• Describe the discharge of a capacitor through a resistor, use the time‑constant τ = RC, and sketch the exponential decay.

1. What is Capacitance?

Capacitance, C, measures a system’s ability to store electric charge per unit potential difference. By definition

C = \dfrac{Q}{V}

where Q is the magnitude of charge on one conductor and V is the potential difference between the two conductors (or between a single isolated conductor and infinity). The SI unit is the farad (F), where 1 F = 1 C V⁻¹.

2. Isolated Spherical Conductor

Consider a conducting sphere of radius R carrying charge Q. In vacuum the electric potential relative to infinity is

V = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Q}{R}

Substituting into the definition of capacitance gives

C{\text{sphere}} = \dfrac{Q}{V}=4\pi\varepsilon{0}R

• Capacitance is directly proportional to the radius.
• Only the geometry (R) and the permittivity of free space \varepsilon_{0}=8.85\times10^{-12}\;{\rm F\,m^{-1}} matter.

Dielectric surrounding the sphere – If the sphere is immersed in a material of permittivity \varepsilon = \kappa\varepsilon_{0}, the formula becomes

C = 4\pi\varepsilon R = 4\pi\kappa\varepsilon_{0}R

Worked Example – Isolated Sphere

1. Radius R = 5.0 cm = 0.050 m.
2. In vacuum: C = 4\pi(8.85\times10^{-12})(0.050) \approx 5.6\times10^{-12}\;F = 5.6 pF.
3. If the sphere is surrounded by a dielectric with \kappa = 2.0, C = 2.0 × 5.6 pF = 11.2 pF.

3. Parallel‑Plate Capacitor

A parallel‑plate capacitor consists of two large, flat plates of area A separated by a small distance d (d \ll \sqrt{A}). The space between the plates may contain a dielectric of relative permittivity \kappa (for vacuum or air, \kappa = 1).

Uniform electric field between the plates:

E = \dfrac{V}{d}

Charge on each plate (using D = \varepsilon_{0}\kappa E and Q = D A):

Q = \varepsilon_{0}\kappa A E

Combining the two expressions gives the familiar formula

C{\text{parallel}} = \dfrac{Q}{V}= \varepsilon{0}\kappa \dfrac{A}{d}

• Capacitance increases with larger plate area A.
• Capacitance decreases when the separation d increases.
• Inserting a dielectric multiplies the capacitance by the factor \kappa (e.g., mica: \kappa ≈ 5–7).

Worked Example – Parallel‑Plate Capacitor

1. A = 0.020 m², d = 1.0 mm = 1.0\times10^{-3}\;m, \kappa = 2.5 (e.g., glass).
2. C = (8.85\times10^{-12})(2.5)\dfrac{0.020}{1.0\times10^{-3}} \approx 4.4\times10^{-10}\;F = 440 pF.

4. Combining Capacitors

4.1 Parallel Combination

All capacitors experience the same voltage. The total capacitance is the algebraic sum:

C{\text{eq}} = C{1}+C{2}+C{3}+ \dots

4.2 Series Combination

All capacitors carry the same charge. The reciprocal of the equivalent capacitance is the sum of reciprocals:

\dfrac{1}{C{\text{eq}}}= \dfrac{1}{C{1}}+\dfrac{1}{C{2}}+\dfrac{1}{C{3}}+\dots

4.3 Mixed (Series‑Parallel) Example – Exam‑style

Three capacitors: C{1}=10 pF, C{2}=20 pF, C{3}=30 pF. C{1} and C{2} are connected in parallel, and this combination is in series with C{3}.

1. Parallel part: C{12}=C{1}+C_{2}=10 pF+20 pF=30 pF.
2. Series with C_{3}:

   \(\displaystyle \frac{1}{C{\text{eq}}}= \frac{1}{C{12}}+\frac{1}{C_{3}}=\frac{1}{30\,\text{pF}}+\frac{1}{30\,\text{pF}}=\frac{2}{30\,\text{pF}}\)

   Thus C_{\text{eq}} = \dfrac{30\,\text{pF}}{2}=15 pF.

5. Energy Stored in a Capacitor

Numeric Example

1. Capacitance C = 100 pF, voltage V = 12 V.
2. \(W = \tfrac{1}{2}(100\times10^{-12})(12)^{2} \approx 7.2\times10^{-9}\;J = 7.2\;nJ.\)

6. Discharging a Capacitor – RC Circuits

When a charged capacitor is connected across a resistor R, the charge and voltage decay exponentially.

• Time constant: \(\tau = RC\) (seconds).
• Voltage decay: \(V(t) = V_{0}\,e^{-t/\tau}\).
• Charge decay: \(Q(t)=Q_{0}\,e^{-t/\tau}\).

Key points:

• After one time constant (\(t=\tau\)), the voltage has fallen to \(\approx 37\%\) of its initial value.
• After \(5\tau\) the voltage is <0.7 % of the initial value – effectively discharged.

Worked Example – RC Discharge

1. Capacitor: \(C = 47\;\mu\text{F}\); resistor: \(R = 10\;k\Omega\).
2. \(\tau = RC = (10\times10^{3})(47\times10^{-6}) = 0.47\;s.\)
3. Initial voltage \(V_{0}=5\;V\). Voltage after \(t = 1\;s\):

   \[

   V(1) = 5\,e^{-1/0.47} \approx 5\,e^{-2.13} \approx 0.6\;V.

   \]

Suggested diagram: A simple RC circuit (capacitor discharging through a resistor) with an over‑laid exponential decay curve.

7. Comparison of the Two Basic Configurations

8. Key Take‑aways

• Capacitance is a geometric property; it does not depend on the amount of charge stored.
• For isolated conductors the reference potential is infinity; for a capacitor it is the opposite plate.
• Series and parallel combination rules are algebraic and essential for circuit design and exam problems.
• Energy stored: \(W = \tfrac{1}{2}CV^{2}\) (or the equivalent forms). The factor ½ reflects the linear increase of voltage during charging.
• During discharge, voltage and charge fall exponentially with time constant \(\tau = RC\); after \(5\tau\) the capacitor is effectively empty.