(a) Explain the statement the capacitance of a capacitor is 5\(\mu\)F. (b)(i) State the factors upon which the capacitance of a parallel plate capacitor dep...

(a) Capacitance is defined by \(C=\dfrac{Q}{V}\). Therefore, a capacitance of \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) on each plate when the potential difference across it is \(1\,\text{V}\).

(b)(i) For a parallel-plate capacitor, capacitance depends on:

• the common area of overlap of the plates, \(A\);
• the separation between the plates, \(d\);
• the permittivity, \(\varepsilon\), of the dielectric between the plates.

(b)(ii) The relationship is

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

• \(C\) is directly proportional to plate area: \(C\propto A\).
• \(C\) is inversely proportional to plate separation: \(C\propto \dfrac{1}{d}\).
• \(C\) is directly proportional to the permittivity of the dielectric: \(C\propto\varepsilon\).

(c)(i) The capacitors are connected one after another in a single series path. The switch is shown open.

(c)(ii) For capacitors in series, the reciprocals of the capacitances are added:

\[ \frac{1}{C_{\text{effective}}} =\frac{1}{8}+\frac{1}{12}+\frac{1}{24}. \] \[ \frac{1}{C_{\text{effective}}} =\frac{3+2+1}{24} =\frac{6}{24} =\frac{1}{4}. \] \[ C_{\text{effective}}=4\,\mu\text{F}. \]

(c)(iii) In a series circuit, each capacitor has the same magnitude of charge. Thus:

\[ Q=C_{\text{effective}}V =(4\,\mu\text{F})(90\,\text{V}) =360\,\mu\text{C}. \]

Therefore, the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\), and \(24\,\mu\text{F}\) capacitors is \(360\,\mu\text{C}\).

(c)(iv) The potential difference across the \(8\,\mu\text{F}\) capacitor is:

\[ V=\frac{Q}{C} =\frac{360\,\mu\text{C}}{8\,\mu\text{F}} =45\,\text{V}. \]

The p.d. across the \(8\,\mu\text{F}\) capacitor is therefore \(45\,\text{V}\). In a series combination, the smallest capacitance has the largest potential difference because \(V=\dfrac{Q}{C}\) and the charge is the same on all capacitors.