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A capacitor of capacitance 63 pF is made from two parallel metal plates separated by an air gap - AQA - A-Level Physics - Question 1 - 2021 - Paper 2
Question 1
A capacitor of capacitance 63 pF is made from two parallel metal plates separated by an air gap. The capacitor is charged so that it stores a charge of 7.6 × 10⁻¹⁰ ... show full transcript
Worked Solution & Example Answer:A capacitor of capacitance 63 pF is made from two parallel metal plates separated by an air gap - AQA - A-Level Physics - Question 1 - 2021 - Paper 2
Step 1
Explain what is meant by a dielectric constant of 6.0
Answer
The dielectric constant, or relative permittivity, of a material indicates how much the material can reduce the electric field compared to a vacuum. A dielectric constant of 6.0 means the permittivity of the material is 6 times that of free space, allowing for greater charge storage and affecting the capacitance of the capacitor.
Step 2
Explain how the polar molecules cause this change in capacitance
Answer
When the mica dielectric is inserted, the polar molecules align themselves in response to the electric field of the capacitor. The positive ends of the polar molecules will face the negative plate of the capacitor, and the negative ends will face the positive plate. This alignment reduces the effective electric field between the plates and increases the capacitance, as it allows the plates to store more charge for the same applied voltage.
Step 3
Calculate the difference between the initial energy stored by the capacitor and the energy stored when the mica has been fully inserted.
Answer
The initial energy stored in the capacitor can be calculated using the formula: E_i = rac{1}{2} C V^2 Where:
- Capacitance, C = 63 pF = 63 × 10⁻¹² F
- Voltage, V can be calculated from charge, Q = 7.6 × 10⁻¹⁰ C using the formula: V = rac{Q}{C} = rac{7.6 imes 10^{-10}}{63 imes 10^{-12}} = 12.06 V
Thus, E_i = rac{1}{2} imes 63 imes 10^{-12} imes (12.06)^2 = 4.58 imes 10^{-10} J
When the mica is fully inserted, the new capacitance is calculated as:
Using the same voltage: E_f = rac{1}{2} C' V^2 = rac{1}{2} imes 378 imes 10^{-12} imes (12.06)^2 = 2.78 imes 10^{-9} J
The difference in energy is therefore:
Step 4
Sketch a graph on Figure 2 to show how the capacitance C varies with θ as the spindle is turned through 360°
Answer
The graph should show a linear increase in capacitance C from 0 (when θ = 0°, plates completely overlapping) to a maximum value (C_max) when θ reaches 360° (fully separated). The x-axis represents the angle θ, and the y-axis represents capacitance C. A straight line from (0, C_max) to (360, C_max) indicates this relationship.
Step 5
Explain, with numerical detail, two ways in which this can be achieved.
Answer
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Halve the spacing between the plates: The capacitance of a parallel plate capacitor can be increased by decreasing the distance d between the plates. The formula for capacitance is given by: C = rac{ ext{ε} A}{d} If the distance is halved, the capacitance will double.
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Increase the area of the plates: Doubling the area A of the plates directly increases the capacitance, as shown in the same formula. Thus, if the diameter of the plates is increased, more charge can be stored.
By either of these methods, the theoretical maximum capacitance can be obtained without altering the original capacitor's fundamental properties.