(a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2\(\mu\)F (b) State: (i) three factors on which its capacitance...

(a) Meaning of \(2\,\mu\text{F}\)

Capacitance is defined by \(C=\dfrac{Q}{V}\). A capacitance of \(2\,\mu\text{F}\) means that the capacitor stores \(2\,\mu\text{C}\) of charge on each plate for every \(1\,\text{V}\) potential difference across its plates:

\[C=2\,\mu\text{F}=2\times10^{-6}\,\text{F}\]

Thus, when \(V=1\,\text{V}\), \(Q=CV=2\times10^{-6}\,\text{C}=2\,\mu\text{C}\). The two plates carry equal charges of opposite sign.

(b)(i) Factors affecting the capacitance of a parallel-plate capacitor

• The area of overlap of the plates: larger area gives greater capacitance.
• The separation between the plates: greater separation gives smaller capacitance.
• The permittivity of the dielectric between the plates: a dielectric with higher permittivity gives greater capacitance.

For parallel plates, this is summarised by:

\[C=\frac{\varepsilon A}{d}\]

(b)(ii) Uses of capacitors

• Storing charge or energy, for example in a camera flash.
• Smoothing the output from a rectifier in a power supply.
• Separating a.c. signals from d.c. components in electronic circuits.

(c) Derivation of the energy stored

During charging, the potential difference is not constant: it rises from \(0\) to its final value \(V\). When the charge already on the capacitor is \(q\),

\[V=\frac{q}{C}\]

The work done in bringing a further small charge \(\mathrm{d}q\) onto the capacitor is:

\[\mathrm{d}W=V\,\mathrm{d}q=\frac{q}{C}\,\mathrm{d}q\]

Therefore, the total energy stored when the final charge is \(Q\) is:

\[ W=\int_0^Q\frac{q}{C}\,\mathrm{d}q =\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q =\frac{Q^2}{2C} \]

Since \(Q=CV\), equivalent forms are:

\[ \boxed{W=\frac{Q^2}{2C}=\frac{1}{2}QV=\frac{1}{2}CV^2} \]

The use of \(\tfrac12 QV\), rather than \(QV\), is important because the p.d. increases gradually while the capacitor is charged.

(d) Capacitors connected in parallel to a \(50\,\text{V}\) d.c. supply

In a parallel connection, each capacitor is connected directly across the source. Therefore, the potential difference across each is \(50\,\text{V}\).

(i) Charge on either plate of each capacitor

For the \(2\,\mu\text{F}\) capacitor:

\[ Q_1=C_1V=(2\times10^{-6})(50) =1.0\times10^{-4}\,\text{C} =100\,\mu\text{C} \]

For the \(3\,\mu\text{F}\) capacitor:

\[ Q_2=C_2V=(3\times10^{-6})(50) =1.5\times10^{-4}\,\text{C} =150\,\mu\text{C} \]

Each capacitor has these charge magnitudes on its plates: one plate is positive and the other has an equal negative charge.

(ii) Potential difference across each capacitor

\[ \boxed{V_1=V_2=50\,\text{V}} \]

(iii) Energy stored by the combined capacitors

For capacitors in parallel, capacitances add:

\[ C_{\text{total}}=2\,\mu\text{F}+3\,\mu\text{F}=5\,\mu\text{F} =5\times10^{-6}\,\text{F} \]

\[ W=\frac12C_{\text{total}}V^2 =\frac12(5\times10^{-6})(50)^2 =6.25\times10^{-3}\,\text{J} \]

\[ \boxed{W=6.25\times10^{-3}\,\text{J}} \]

Examination reminder: In parallel, the potential difference is the same across every capacitor, while the charge on each capacitor is found separately using \(Q=CV\).