All insulated metal bodies can store charge - Leaving Cert Physics - Question 7 - 2020
Question 7
All insulated metal bodies can store charge.
(i) Describe how a pear-shaped metal body can be charged by induction.
(ii) Draw a diagram to show the distribution of... show full transcript
Worked Solution & Example Answer:All insulated metal bodies can store charge - Leaving Cert Physics - Question 7 - 2020
Step 1
Describe how a pear-shaped metal body can be charged by induction.
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Answer
A pear-shaped metal body can be charged by induction through the following process:
Bring Charged Body Close: Place a charged object near the pear-shaped conductor, which causes the charges within the conductor to redistribute. The side closest to the charged object will undergo a slight negative charge, while the far side becomes positively charged.
Connect the Conductor to Earth: Connect the conductor to the ground to allow the excess positive charge to escape to the earth, neutralizing that side of the conductor.
Remove the Connection to Earth: Disconnect the connection to the earth while the charged object is still nearby. This retains the negative charge on the conductor, leaving it electrically charged.
Step 2
Draw a diagram to show the distribution of charge on the body after charging.
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Answer
A diagram illustrating the pear-shaped conductor would show.
The charged object nearby.
Negative charges concentrated at the nearest point to the charged object and positive charges distributed on the far side.
Step 3
Define capacitance.
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Answer
Capacitance (C) is defined as the ability of a system to store an electric charge per unit voltage. The formula for capacitance is given by:
C=VQ
where Q is the charge stored and V is the voltage across the capacitor.
Step 4
Calculate the maximum rise in temperature of the water.
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Answer
The energy stored in the capacitor (E) is given by:
E=21CV2
Substituting the given values:
E=21×4000×10−6F×(500V)2=500J
Using the specific heat formula:
Q=mcΔT
We have,
mass (m) = 0.04 kg,
specific heat capacity (c) = 4180 J kg⁻¹ K⁻¹,
energy (Q) = 500 J.
Rearranging for the rise in temperature (ΔT):
ΔT=mcQ=0.04kg×4180Jkg−1K−1500J=2.98K
Step 5
Describe an experiment to demonstrate how the capacitance of a parallel-plate capacitor changes with the distance between the plates.
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Answer
Apparatus: Use a parallel-plate capacitor, a power supply, and a capacitance meter.
Method: Measure the capacitance at varying distances by adjusting the separation of the plates.
Observations: Record the capacitance values for each distance and graph them. As the distance between the plates increases, the capacitance decreases, demonstrating the inverse relationship.
Step 6
Calculate the surface area of the inner cylinder of aluminium foil.
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Answer
To find the surface area (A) of the inner cylinder:
A=2πrh
where r=0.06m and h=0.17m.
Calculating:
A=2π×0.06m×0.17m=0.064m2
Step 7
Calculate the capacitance of the Leyden jar.
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Answer
The capacitance (C) can be calculated using the formula:
C=ε0dA
where:
ε0 = 8.85 × 10⁻¹² F/m,
A is the surface area calculated earlier,
d is the dielectric thickness (0.005 m).
Continuing from the previous calculation, substituting values will yield:
C≈44µF
Step 8
What property of glass allows it to be used as a dielectric?
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Answer
The property of glass that allows it to be used as a dielectric is its high dielectric strength and its relative permittivity, which increases the capacitor's ability to store charge without conducting electricity.
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