(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) 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 depend.
(ii) State how the capacitance depends on each of these factors stated in (b)(i).
(c) A series arrangement of three capacitors of values 8uF, 12\(\mu\)F, and 24\(\mu\)F is connected in series with 90-V battery.
(i) Draw an open-circuit diagram for this arrangement.
(ii) Calculate the effective capacitance in the circuit.
(iii) On closed circuit, calculate the charge on each capacitor when fully charged.
(iv) Determine the p.d across the 8\(\mu\)F capacitor.
(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:
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.
(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:
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.
