(ii) A parallel plate capacitor consists of two plates each of area 9.6 x 10\(^{-2}\)m\(^2\), separated by a dielectric of thickness 2.25 x 10\(^{-3}\) and dielectric constant 900. Calculate the capacitance of the capacitor. [ \(\varepsilon_o\) = permittivity of free space = 8.85 x 10\(^{-12}\)Fm\(^{-1}\)]
(b) (i) Which of the following devices has a higher resistance; an ammeter or a voltmeter? Give a reason for your answer.
(a)(i) Dielectric
A dielectric is an insulating material (solid, liquid or gas) placed between the two plates of a capacitor. When placed in the field it becomes polarised and increases the capacitance of the capacitor. Examples are mica, waxed paper and air.
(a)(ii) Capacitance
\[ C = \frac{\varepsilon_r\,\varepsilon_0\,A}{d} \]
with \(\varepsilon_r = 900\), \(\varepsilon_0 = 8.85\times10^{-12}\,\text{Fm}^{-1}\), \(A = 9.6\times10^{-2}\,\text{m}^2\) and \(d = 2.25\times10^{-3}\,\text{m}\):
\[ C = \frac{900 \times 8.85\times10^{-12} \times 9.6\times10^{-2}}{2.25\times10^{-3}} = 3.4\times10^{-7}\,\text{F} \]
(b)(i) Ammeter or voltmeter?
The voltmeter has the higher resistance. A voltmeter is connected in parallel with the component whose p.d. is being measured, so it must have a very high resistance to draw only a negligible current and avoid disturbing the circuit. An ammeter is connected in series and must have a very low resistance so that it does not alter the current it measures.
(b)(ii) Voltmeter reading
The two 400 \(\Omega\) resistors on the left are in parallel:
\[ \frac{1}{R_1} = \frac{1}{400} + \frac{1}{400} \Rightarrow R_1 = 200\,\Omega \]
The 800 \(\Omega\) resistor is in parallel with the 800 \(\Omega\) voltmeter:
\[ \frac{1}{R_2} = \frac{1}{800} + \frac{1}{800} \Rightarrow R_2 = 400\,\Omega \]
\(R_1\) and \(R_2\) are in series across the 6 V supply:
\[ R_T = 200 + 400 = 600\,\Omega, \qquad I = \frac{V_T}{R_T} = \frac{6}{600} = 0.01\,\text{A} \]
The voltmeter reads the p.d. across \(R_2\):
\[ V = I\,R_2 = 0.01 \times 400 = 4.0\,\text{V} \]
(c) Equivalent resistance
The two 2 \(\Omega\) resistors in parallel combine to:
\[ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow R_p = 1\,\Omega \]
This 1 \(\Omega\) is in series with the remaining 2 \(\Omega\) and 2 \(\Omega\) resistors:
\[ R_T = 2 + 1 + 2 = 5\,\Omega \]
