PHY_10_37_V2_DM_Capacitors

Problem Solving on topic Capacitors. Capacitance.

1.    A parallel plate capacitor with a plate separation of 1.0mm1.0mm has an electric field of magnitude 2.4×104Vm−12.4×104Vm−1 inside it. What is the potential difference across this capacitor?

Answer: 24V

2.    A parallel plate capacitor has an area of 4.0mm24.0mm2 and an air gap of 1.0mm1.0mm. What is its capacitance?

Answer: 0.035pF

3.    A dielectric material with 𝜀=1.2×10−10Fm−1ε=1.2×10−10Fm−1 is placed into the same parallel plate capacitor (area of 4.0mm24.0mm2, gap of 1.0mm1.0mm). What is its new capacitance?

Answer: 0.48pF

4.    This capacitor (area of 4.0mm24.0mm2, gap of 1.0mm1.0mm containing a material with 𝜀=1.2×10−10Fm−1ε=1.2×10−10Fm−1) now holds a charge of 3.6×10−11C3.6×10−11C. What is the potential difference across it?

Answer: 75V

5.     An electron of mass 9.11 ´ 10–31 kg is at rest on the negative plate of a parallel plate capacitor. The electron is repelled by the negative plate and attracted to the positive plate by a force of 6.45 ´ 10–16 N. The distance between the plates is 2.50 mm. The charge on 1 electron is –1.60 ´ 10–19 C. a) Sketch the electric field between the plates of the charged parallel plate capacitor.

Answer: (2 marks) The electric field pattern should resemble the one below.

b) Calculate the electric field between the plates of the charged parallel plate capacitor.

 Answer: (2 marks) Given: Separation of plates d = 2.50 mm = 2.50 ´ 10–3 m Charge on electron q = -1.60 ´ 10–19 C Electric force = 6.45 ´ 10–16 N Mass of electron mE = 9.11 ´ 10–31 kg Unknown: Electric field

The electric field between the plates is 4030 N/C towards the negative plate.

c) What is the velocity of the electron just before it crashes into the positive plate?

Answer: (3 marks) The work done by the electric force will become the kinetic energy of the electron. Equation: 

The electron hits the positive plate travelling at 1.88 x10⁶ m/s.

Hangouts for students:

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