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SS3: PHYSICS - 1ST TERM

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  1. Energy & Society
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  2. Electromagnetic Waves
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  3. Gravitational Field
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  4. Electric Field I
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SS3: PHYSICS – 1ST TERM Capacitance & Capacitor Theory Questions – Capacitance & Capacitor
Lesson 5, Topic 5
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Theory Questions – Capacitance & Capacitor

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Topic Content:

  • Theory Questions & Answers - Capacitance & Capacitor:

Theory Questions - Capacitance & Capacitor:

1. (a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2 μ F

(b) State:

(i) three factors on which its capacitance depends

(ii) three uses of capacitors.

(c) Derive a formula for the 

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Theory Questions & Answers – Capacitance & Capacitor

1. (a) Explain what is meant by the statement: The capacitance of a parallel-plate capacitor is 2 μF

Answer:

The capacitance of the parallel plate capacitor determines the amount of charge that it can hold. The statement means that the ratio of the charge on the capacitor to the potential difference between the plates is 2 × 10-6 F or 2 micro Farad.

For a 2 micro Farad capacitance, if the voltage increases by 1 V, the charge increases by 2 micro-coulombs.

 

(b) State:

(i) Three factors on which its capacitance depends

Answer:

  • Area of plates (A)
  • Dielectric between plates (ε)
  • Distance between plates (d)

 

(ii) three uses of capacitors.

Answer:

  • They are used in tuning radio and television receivers to the desired stations.
  • They are used as rectifiers of current in a.c source.
  • They prevent sparks in switches like ignition relays.
  • They are used as storage of large amounts of charges in research institutes.
  • They are used as timers.

 

(c) Derive a formula for the energy W stored in a charged capacitor of capacitance C carrying a charge Q on either plate.

A capacitor is a two-terminal electrical device that can store energy in the form of an electric charge. Initially, the charge on the plates is q = 0. As the capacitor is being charged, the charge gradually builds up on its plates, and after some time, it reaches the value q.

When charging a capacitor, the average value of the potential difference is given by:

⇒ \( \frac{0 + V}{2} \)

= \( \frac{1}{2} \scriptsize V \)

The work done in charging the capacitor through an average potential difference \( \frac{1}{2} \scriptsize V \) is

W = \( \frac{1}{2} \scriptsize V \; \times \; q \)

W =  \( \frac{1}{2} \scriptsize qV \)

But C = \( \frac{q}{V} \)

or V = \( \frac{q}{C} \)

W =  \( \frac{1}{2} \scriptsize qV = \normalsize \frac{1}{2} \scriptsize q \: \times\: \normalsize \frac{q}{C} \)

W = \( \frac{1}{2} \frac{q^2}{C} \)

But q = CV

∴ W = \( \frac{1}{2} \frac{(CV)^2}{C} \)

∴ W = \( \frac{1}{2} \frac{C^2V^2}{C} \)

∴ W = \( \frac{1}{2} \frac{\not{C}^2V^2}{\not{C}} \)

W =  \( \frac{1}{2} \scriptsize CV^2 \)

∴ W = \(\normalsize \frac{1}{2} \scriptsize qV = \normalsize \frac{1}{2} \frac{q^2}{C} = \normalsize \frac{1}{2} \scriptsize CV^2 \)

 

(d) Two parallel-plate capacitors of capacitances 2 μF and 3 μF are connected in parallel and the combination is connected to a 50 V d.c. source. Draw the circuit diagram of the arrangement and determine the:

(i) charge on either plate of each capacitor

(ii) potential difference across each capacitor

(iii) energy of the combined capacitors

Solution

(i) Using:

C = \( \frac{Q}{V} \)

∴ Q = CV

Charge on either plate of C1:

Q1 = C1V

Q1 = 2 × 10-6 × 50 = 1 ×  10-4  C or 100 μC

Charge on either plate of C2:

Q2 = C2V

Q2 = 3 × 10-6 × 50 = 1.5 ×  10-4  C or 150 μC

 

(ii) potential difference across each capacitor

Answer: P.d across each capacitor is 50V

 

(iii) energy of the combined capacitors

Solution

Capacitors are in parallel

∴ Total capacitance, CT = C1 + C2

CT = 2 μ F + 3 μ F = 5  μ F

W =  \( \frac{1}{2} \scriptsize C_TV^2 \)

W =  \( \frac{1}{2}\scriptsize \: \times \: 5 \: \times \: 10^{-6} \: \times \: 50 \: \times \: 50  \)

W =  \( \frac{1}{2}\scriptsize \: \times \: 0.0125  \)

W =  \( \scriptsize  6.25\: \times \: 10^{-3} \: Joules  \)

Theory Questions & Answers – Capacitance & Capacitor

2. The plates of a parallel plate capacitor in vacuum are 5 mm apart and 2 m2 in area. A potential difference of 10000 V is applied across the capacitor. Find the

(a) capacitance

(b) the charge on each plate

(c) the magnitude of the electric field between the plates?

(d) Energy stored in the capacitor

Solution

(a) capacitance

To calculate the capacitance (C) of a parallel-plate capacitor, we can use the formula:

C = \( \frac{Aε_0}{d} \)

where:

  • C is the capacitance
  • ε0 is the permittivity of free space (vacuum), approximately 8.85 × 10-12 F/m
  • A is the area of each plate
  • d is the separation between the plates

Given:

A = 2.00 m2
d = 5.00 mm = 5.00 × 10-3 m

Using these values, we can calculate the capacitance:

C = \( \frac{2 \: \times \: 8.85 \: \times \: 10^{-12}}{5 \: \times \: 10^{-3}} \)

C = \( \scriptsize 3.54 \: \times \: 10^{-9} \: F\)

 

(b) the charge on each plate

Q = CV

  • V = 10000 V
  • C = \( \scriptsize 3.54 \: \times \: 10^{-9} \: F\)

Q = \( \scriptsize 3.54 \: \times \: 10^{-9} \: \times \: 10000\)

Q = \( \scriptsize 3.54 \: \times \: 10^{-5} \:C\)

 

(c) the magnitude of the electric field between the plates?

E = \( \frac{V}{d}\)

where:
E is the magnitude of the electric field
V is the potential difference applied across the capacitor
d is the separation between the plates

Using the given values, we can calculate the magnitude of the electric field:

E = \( \frac{10000}{5 \: \times \: 10^{-3} }\)

E = \( \scriptsize 2 \: \times \: 10^{-6} \: Vm^{-1}\)

 

(d) Energy stored in the capacitor

W = \(\frac{1}{2} \scriptsize CV^2 \)

W = \(\frac{1}{2} \scriptsize \: \times \: 3.54 \: \times \: 10^{-9} \: \times \: 10000^2 \)

W = \( \scriptsize 0.177 \: J \: or \: 177 mJ\)

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