All insulated metal bodies can store charge - Leaving Cert Physics - Question 7 - 2020

To illustrate the charge distribution after charging, draw a pear-shaped outline. Indicate with '+' signs a concentration of positive charge at the pointed end and '-' signs at the wider end, representing a negative charge accumulation if the original charged body was positive.

Capacitance is defined as the ability of a capacitor to store electrical charge. Mathematically, it is represented by the formula: $C = \frac{Q}{V}$ where:

• $C$ is the capacitance in farads (F).
• $Q$ is the charge stored in coulombs (C).
• $V$ is the voltage across the capacitor in volts (V).

The energy stored in the capacitor is given by the formula: $E = \frac{1}{2} C V^2$ Given:

• Capacitance, $C = 4000 \mu F = 4000 \times 10^{-6} F$
• Voltage, $V = 500 V$

Calculating the energy:

$E = \frac{1}{2} \times 4000 \times 10^{-6} \times (500)^2 = 500 J$

The heat transferred to the water can be calculated using: $Q = mc \Delta T$ where:

• $m = 40 g = 0.04 kg$ (mass of water)
• $c = 4180 J kg^{-1} K^{-1}$ (specific heat capacity of water)
• $heta$ is the temperature change.

Substituting the values:

$500 = 0.04 \times 4180 \times \Delta T$

Solving for $heta$: $\Delta T = \frac{500}{0.04 \times 4180} = 2.98 K$

Therefore, the maximum rise in temperature of the water is approximately 3 °C.

To demonstrate how the capacitance of a parallel-plate capacitor changes with plate separation, conduct the following experiment:

Apparatus:

• A parallel-plate capacitor with adjustable distance between the plates.
• A multimeter to measure capacitance.
• A ruler to measure the distance between the plates.

Method:

1. Set up the parallel-plate capacitor.
2. Use the ruler to adjust the distance between the plates to different predetermined values.
3. Measure the capacitance using the multimeter for each distance setting.

Observations: Record the capacitance readings as the distance increases. A graph can be plotted of capacitance (C) versus distance (d). According to the principle, capacitance decreases as the distance between the plates increases.

The surface area $A$ of the inner cylinder can be calculated using the formula: $A = 2 \pi r h$ Where:

• $r = 0.06 m$ (internal radius, converting 6 cm to meters)
• $h = 0.17 m$ (height of the aluminium foil)

Substituting the values: $A = 2 \pi \times 0.06 \times 0.17 = 0.0638 m^2$ The surface area of the inner cylinder of aluminium foil is approximately 0.0638 m².

The capacitance $C$ of a Leyden jar can be calculated using: $C = \varepsilon_0 \varepsilon_r \frac{A}{d}$ Where:

• $\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$ (vacuum permittivity)
• $\varepsilon_r = 2.1$ (relative permittivity of the glass)
• $A = 0.0638 m^2$ (from previous calculation)
• $d = 0.005 m$ (thickness of the glass)

Substituting the values: $C = 8.85 \times 10^{-12} \times 2.1 \times \frac{0.0638}{0.005} = 2.38 \times 10^{-8} F = 23.8 nF$ Therefore, the capacitance of the Leyden jar is approximately 23.8 nF.

Glass is used as a dielectric due to its insulating properties. It has a high dielectric strength, which allows it to withstand high electric fields without conducting electricity. The relative permittivity of glass enhances the capacitor's ability to store charge by reducing the electric field between the plates, thereby increasing capacitance.