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

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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:

  1. 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.
  2. 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.
  3. 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.

  1. The charged object nearby.
  2. 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=QVC = \frac{Q}{V} where QQ is the charge stored and VV 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=12CV2E = \frac{1}{2} C V^2 Substituting the given values: E=12×4000×106F×(500V)2=500JE = \frac{1}{2} \times 4000 \times 10^{-6} F \times (500 V)^2 = 500 J

Using the specific heat formula:

Q=mcΔTQ = mc \Delta T We have,

  • mass (mm) = 0.04 kg,
  • specific heat capacity (cc) = 4180 J kg⁻¹ K⁻¹,
  • energy (QQ) = 500 J.

Rearranging for the rise in temperature (ΔT\Delta T):

ΔT=Qmc=500J0.04kg×4180Jkg1K1=2.98K\Delta T = \frac{Q}{mc} = \frac{500 J}{0.04 kg \times 4180 J kg^{-1} K^{-1}} = 2.98 K

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

  1. Apparatus: Use a parallel-plate capacitor, a power supply, and a capacitance meter.
  2. Method: Measure the capacitance at varying distances by adjusting the separation of the plates.
  3. 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=2πrhA = 2 \pi r h where r=0.06mr = 0.06 m and h=0.17mh = 0.17 m. Calculating: A=2π×0.06m×0.17m=0.064m2A = 2 \pi \times 0.06 m \times 0.17 m = 0.064 m^2

Step 7

Calculate the capacitance of the Leyden jar.

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Answer

The capacitance (C) can be calculated using the formula:

C=ε0AdC = \varepsilon_0 \frac{A}{d} where:

  • ε0\varepsilon_0 = 8.85 × 10⁻¹² F/m,
  • AA is the surface area calculated earlier,
  • dd is the dielectric thickness (0.005 m). Continuing from the previous calculation, substituting values will yield: C44µFC ≈ 44 µF

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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