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Capacitance and the parallel plate capacitor

Capacitors and capacitance

A parallel plate capacitor is a device that consists of two oppositely charged conducting plates separated by a small distance, which stores charge. When voltage is applied to the capacitor, usually by connecting it to an energy source (e.g. a battery) in a circuit, electric charge of equal magnitude, but opposite polarity, builds up on each plate.

Figure 1: A capacitor (C) connected in series with a resistor (R) and an energy source (E).
Figure 1 (PG11C7_023.png)
Definition 1: Capacitance

Capacitance is the charge stored per volt and is measured in Farads (F).

Mathematically, capacitance is the ratio of the charge on a single plate to the voltage across the plates of the capacitor:

C = Q V . C = Q V .
(1)

Capacitance is measured in Farads (F). Since capacitance is defined as C=QVC=QV, the units are in terms of charge over potential difference. The unit of charge is the coulomb and the unit of the potential difference is the volt. One farad is therefore the capacitance if one coulomb of charge was stored on a capacitor for every volt applied.

1 C of charge is a very large amount of charge. So, for a small amount of voltage applied, a 1 F capacitor can store a enormous amount of charge. Therefore, capacitors are often denoted in terms of microfarads (1×10-61×10-6), nanofarads (1×10-91×10-9), or picofarads (1×10-121×10-12).

Tip:

QQ is the magnitude of the charge stored on either plate, not on both plates added together. Since one plate stores positive charge and the other stores negative charge, the total charge on the two plates is zero.

Exercise 1: Capacitance

Suppose that a 5 V battery is connected in a circuit to a 5 pF capacitor. After the battery has been connected for a long time, what is the charge stored on each of the plates?

Solution
  1. Step 1. Determine what we know :

    To begin remember that after a voltage has been applied for a long time the capacitor is fully charged. The relation between voltage and the maximum charge of a capacitor is found in equation  (Reference).

    C V = Q C V = Q
    (2)

    Inserting the given values of C=5FC=5F and V=5VV=5V, we find that:

    Q = C V = ( 5 × 10 - 12 F ) ( 5 V ) = 2 , 5 × 10 - 11 C Q = C V = ( 5 × 10 - 12 F ) ( 5 V ) = 2 , 5 × 10 - 11 C
    (3)

Dielectrics

The electric field between the plates of a capacitor is affected by the substance between them. The substance between the plates is called a dielectric. Common substances used as dielectrics are mica, perspex, air, paper and glass.

When a dielectric is inserted between the plates of a parallel plate capacitor the dielectric becomes polarised so an electric field is induced in the dielectric that opposes the field between the plates. When the two electric fields are superposed, the new field between the plates becomes smaller. Thus the voltage between the plates decreases so the capacitance increases.

In every capacitor, the dielectric stops the charge on one plate from travelling to the other plate. However, each capacitor is different in how much charge it allows to build up on the electrodes per voltage applied. When scientists started studying capacitors they discovered the property that the voltage applied to the capacitor was proportional to the maximum charge that would accumulate on the electrodes. The constant that made this relation into an equation was called the capacitance, C. The capacitance was different for different capacitors. But, it stayed constant no matter how much voltage was applied. So, it predicts how much charge will be stored on a capacitor when different voltages are applied.

Physical properties of the capacitor and capacitance

The capacitance of a capacitor is proportional to the surface area of the conducting plate and inversely proportional to the distance between the plates. It also depends on the dielectric between the plates. We say that it is proportional to the permittivity of the dielectric. The dielectric is the non-conducting substance that separates the plates. As mentioned before, dielectrics can be air, paper, mica, perspex or glass.

The capacitance of a parallel-plate capacitor is given by:

C = ϵ 0 A d C = ϵ 0 A d
(4)

where ϵ0ϵ0 is, in this case, the permittivity of air, AA is the area of the plates and dd is the distance between the plates.

Exercise 2: Capacitance

What is the capacitance of a capacitor in which the dielectric is air, the area of the plates is 0,001m20,001m2 and the distance between the plates is 0,02m0,02m?

Solution
  1. Step 1. Determine what is required :

    We need to determine the capacitance of the capacitor.

  2. Step 2. Determine how to approach the problem :

    We can use the formula:

    C = ϵ 0 A d C = ϵ 0 A d
    (5)
  3. Step 3. Determine what is given. :

    We are given the area of the plates, the distance between the plates and that the dielectric is air.

  4. Step 4. Determine the capacitance :
    C = ϵ 0 A d = ( 8 , 9 × 10 - 12 ) ( 0 , 001 ) 0 , 02 = 4 , 45 × 10 - 13 F C = ϵ 0 A d = ( 8 , 9 × 10 - 12 ) ( 0 , 001 ) 0 , 02 = 4 , 45 × 10 - 13 F
    (6)

Electric field in a capacitor

The electric field strength between the plates of a capacitor can be calculated using the formula:

E = V d E = V d

where EE is the electric field in J.C-1J.C-1, VV is the potential difference in volts (VV) and dd is the distance between the plates in metres (mm).

Exercise 3: Electric field in a capacitor

What is the strength of the electric field in a capacitor which has a potential difference of 300V300V between its parallel plates that are 0,02m0,02m apart?

Solution
  1. Step 1. Determine what is required :

    We need to determine the electric field between the plates of the capacitor.

  2. Step 2. Determine how to approach the problem :

    We can use the formula:

    E = V d E = V d

  3. Step 3. Determine what is given. :

    We are given the potential difference and the distance between the plates.

  4. Step 4. Determine the electric field :
    E = V d = 300 0 , 02 = 1 , 50 × 10 4 J . C - 1 E = V d = 300 0 , 02 = 1 , 50 × 10 4 J . C - 1
    (7)

Capacitance and the parallel plate capacitor

  1. Determine the capacitance of a capacitor which stores 9×10-9C9×10-9C when a potential difference of 12 V is applied to it.
  2. What charge will be stored on a 5μF5μF capacitor if a potential difference of 6V6V is maintained between its plates?
  3. What is the capacitance of a capacitor that uses air as its dielectric if it has an area of 0,004m20,004m2 and a distance of 0,03m0,03m between its plates?
  4. What is the strength of the electric field between the plates of a charged capacitor if the plates are 2 mm 2 mm apart and have a potential difference of 200V200V across them?

Figure 2
Figure 2 (capacitor-lab-screenshot.png)
run demo

A capacitor as a circuit device

A capacitor in a circuit

When a capacitor is connected in a DC circuit, current will flow until the capacitor is fully charged. After that, no further current will flow. If the charged capacitor is connected to another circuit with no source of emf in it, the capacitor will discharge through the circuit, creating a potential difference for a short time. This is useful, for example, in a camera flash.

Initially, the electrodes have no net charge. A voltage source is applied to charge a capacitor. The voltage source creates an electric field, causing the electrons to move. The charges move around the circuit stopping at the left electrode. Here they are unable to travel across the dielectric, since electrons cannot travel through an insulator. The charge begins to accumulate, and an electric field forms pointing from the left electrode to the right electrode. This is the opposite direction of the electric field created by the voltage source. When this electric field is equal to the electric field created by the voltage source, the electrons stop moving. The capacitor is then fully charged, with a positive charge on the left electrode and a negative charge on the right electrode.

If the voltage is removed, the capacitor will discharge. The electrons begin to move because in the absence of the voltage source, there is now a net electric field. This field is due to the imbalance of charge on the electrodes–the field across the dielectric. Just as the electrons flowed to the positive electrode when the capacitor was being charged, during discharge, the electrons flow to negative electrode. The charges cancel, and there is no longer an electric field across the dielectric.

Real-world applications: capacitors

Capacitors are used in many different types of circuitry. In car speakers, capacitors are often used to aid the power supply when the speakers require more power than the car battery can provide. Capacitors are also used in processing electronic signals in circuits, such as smoothing voltage spikes due to inconsistent voltage sources. This is important for protecting sensitive electronic components in a circuit.

Figure 3

Summary

  1. Objects can be positively, negatively charged or neutral.
  2. Charged objects feel a force with a magnitude. This is known as Coulomb's law:
    F=kQ1Q2r2F=kQ1Q2r2
    (8)
  3. The electric field due to a point charge is given by the equation:
    E=kQr2E=kQr2
    (9)
  4. The force is attractive for unlike charges and repulsive for like charges.
  5. Electric fields start on positive charges and end on negative charges.
  6. A charge in an electric field, just like a mass under gravity, has potential energy which is related to the work to move it.
  7. A capacitor is a device that stores charge in a circuit.
  8. The electrical potential energy between two point charges is given by:
    U=kQ1Q2r2U=kQ1Q2r2
    (10)
  9. Potential difference is measured in volts and is given by the equation:
    V=WqV=Wq
    (11)
  10. The electric field is constant between equally charged parallel plates. The electric field is given by:
    E=VdE=Vd
    (12)
  11. The capacitance of a capacitor can be calculated as
    C=QV=ϵ0AdC=QV=ϵ0Ad
    (13)

Exercises - Electrostatics

  1. Two charges of +3 nC +3 nC and -5 nC -5 nC are separated by a distance of 40 cm 40 cm . What is the electrostatic force between the two charges?
  2. Two insulated metal spheres carrying charges of +6 nC +6 nC and -10 nC -10 nC are separated by a distance of 20 mm.
    1. What is the electrostatic force between the spheres?
    2. The two spheres are touched and then separated by a distance of 60 mm 60 mm . What are the new charges on the spheres?
    3. What is new electrostatic force between the spheres at this distance?
  3. The electrostatic force between two charged spheres of +3 nC +3 nC and +4 nC +4 nC respectively is 0,04N0,04N. What is the distance between the spheres?
  4. Calculate the potential difference between two parallel plates if it takes 5000J5000J of energy to move 25C25C of charge between the plates?
  5. Draw the electric field pattern lines between:
    1. two equal positive point charges.
    2. two equal negative point charges.
  6. Calculate the electric field between the plates of a capacitor if the plates are 20 mm 20 mm apart and the potential difference between the plates is 300V300V.
  7. Calculate the electrical potential energy of a 6 nC 6 nC charge that is 20 cm 20 cm from a 10 nC 10 nC charge.
  8. What is the capacitance of a capacitor if it has a charge of 0,02C0,02C on each of its plates when the potential difference between the plates is 12V12V?
  9. [SC 2003/11] Two small identical metal spheres, on insulated stands, carry charges -qq and +3q+3q respectively. When the centres of the spheres are separated by a distance dd the one exerts an electrostatic force of magnitude FF on the other.
    Figure 4
    Figure 4 (PG11C7_024.png)
    The spheres are now made to touch each other and are then brought back to the same distance dd apart. What will be the magnitude of the electrostatic force which one sphere now exerts on the other?
    1. 14F14F
    2. 13F13F
    3. 12F12F
    4. 3F3F
  10. [SC 2003/11] Three point charges of magnitude +1 μμC, +1 μμC and -1 μμC respectively are placed on the three corners of an equilateral triangle as shown.
    Figure 5
    Figure 5 (PG11C7_025.png)
    Which vector best represents the direction of the resultant force acting on the -1 μμC charge as a result of the forces exerted by the other two charges?
    Table 1
    Figure 6
    Figure 6 (PG11C7_026.png)
    Figure 7
    Figure 7 (PG11C7_027.png)
    Figure 8
    Figure 8 (PG11C7_028.png)
    Figure 9
    Figure 9 (PG11C7_029.png)
    (a)(b)(c)(d)
  11. [IEB 2003/11 HG1 - Force Fields]
    1. Write a statement of Coulomb's law.
    2. Calculate the magnitude of the force exerted by a point charge of +2 nC on another point charge of -3 nC separated by a distance of 60 mm.
    3. Sketch the electric field between two point charges of +2 nC and -3 nC, respectively, placed 60 mm apart from each other.
  12. [IEB 2003/11 HG1 - Electrostatic Ping-Pong] Two charged parallel metal plates, X and Y, separated by a distance of 60 mm, are connected to a DC supply of emf 2 000 V in series with a microammeter. An initially uncharged conducting sphere (a graphite-coated ping pong ball) is suspended from an insulating thread between the metal plates as shown in the diagram.
    Figure 10
    Figure 10 (PG11C7_030.png)
    When the ping pong ball is moved to the right to touch the positive plate, it acquires a charge of +9 nC. It is then released. The ball swings to and fro between the two plates, touching each plate in turn.
    1. How many electrons have been removed from the ball when it acquires a charge of +9 nC?
    2. Explain why a current is established in the circuit.
    3. Determine the current if the ball takes 0,25 s to swing from Y to X.
    4. Using the same graphite-coated ping pong ball, and the same two metal plates, give TWO ways in which this current could be increased.
    5. Sketch the electric field between the plates X and Y.
    6. How does the electric force exerted on the ball change as it moves from Y to X?
  13. [IEB 2005/11 HG] A positive charge QQ is released from rest at the centre of a uniform electric field.
    Figure 11
    Figure 11 (PG11C7_031.png)
    How does QQ move immediately after it is released?
    1. It accelerates uniformly.
    2. It moves with an increasing acceleration.
    3. It moves with constant speed.
    4. It remains at rest in its initial position.
  14. [SC 2002/03 HG1] The sketch below shows two sets of parallel plates which are connected together. A potential difference of 200 V is applied across both sets of plates. The distances between the two sets of plates are 20 mm and 40 mm respectively.
    Figure 12
    Figure 12 (PG11C7_032.png)
    When a charged particle Q is placed at point R, it experiences a force of magnitude FF. Q is now moved to point P, halfway between plates AB and CD. Q now experiences a force of magnitude .
    1. 12F12F
    2. FF
    3. 2F2F
    4. 4F4F
  15. [SC 2002/03 HG1] The electric field strength at a distance xx from a point charge is EE. What is the magnitude of the electric field strength at a distance 2x2x away from the point charge?
    1. 14E14E
    2. 12E12E
    3. 2E2E
    4. 4E4E
  16. [IEB 2005/11 HG1] An electron (mass 9,11 ×× 10-31-31 kg) travels horizontally in a vacuum. It enters the shaded regions between two horizontal metal plates as shown in the diagram below.
    Figure 13
    Figure 13 (PG11C7_033.png)
    A potential difference of 400 V is applied across the places which are separated by 8 mm. The electric field intensity in the shaded region between the metal plates is uniform. Outside this region, it is zero.
    1. Explain what is meant by the phrase “the electric field intensity is uniform”.
    2. Copy the diagram and draw the following:
      1. The electric field between the metal plates.
      2. An arrow showing the direction of the electrostatic force on the electron when it is at P.
    3. Determine the magnitude of the electric field intensity between the metal plates.
    4. Calculate the magnitude of the electrical force on the electron during its passage through the electric field between the plates.
    5. Calculate the magnitude of the acceleration of the electron (due to the electrical force on it) during its passage through the electric field between the plates.
    6. After the electron has passed through the electric field between these plates, it collides with phosphorescent paint on a TV screen and this causes the paint to glow. What energy transfer takes place during this collision?
  17. [IEB 2004/11 HG1] A positively-charged particle is placed in a uniform electric field. Which of the following pairs of statements correctly describes the potential energy of the charge, and the force which the charge experiences in this field? Potential energy — Force
    1. Greatest near the negative plate — Same everywhere in the field
    2. Greatest near the negative plate — Greatest near the positive and negative plates
    3. Greatest near the positive plate — Greatest near the positive and negative plates
    4. Greatest near the positive plate — Same everywhere in the field
  18. [IEB 2004/11 HG1 - TV Tube] A speck of dust is attracted to a TV screen. The screen is negatively charged, because this is where the electron beam strikes it. The speck of dust is neutral.
    1. What is the name of the electrostatic process which causes dust to be attracted to a TV screen?
    2. Explain why a neutral speck of dust is attracted to the negatively-charged TV screen?
    3. Inside the TV tube, electrons are accelerated through a uniform electric field. Determine the magnitude of the electric force exerted on an electron when it accelerates through a potential difference of 2 000 V over a distance of 50 mm.
    4. How much kinetic energy (in J) does one electron gain while it accelerates over this distance?
    5. The TV tube has a power rating of 300 W. Estimate the maximum number of electrons striking the screen per second.
  19. [IEB 2003/11 HG1] A point charge is held stationary between two charged parallel plates that are separated by a distance d. The point charge experiences an electrical force F due to the electric field E between the parallel plates. What is the electrical force on the point charge when the plate separation is increased to 2d?
    1. 1414 F
    2. 1212 F
    3. 2 F
    4. 4 F
  20. [IEB 2001/11 HG1] - Parallel Plates A distance of 32 mm separates the horizontal parallel plates A and B. B is at a potential of +1 000 V.
    Figure 14
    Figure 14 (PG11C7_034.png)
    1. Draw a sketch to show the electric field lines between the plates A and B.
    2. Calculate the magnitude of the electric field intensity (strength) between the plates. A tiny charged particle is stationary at S, 8 mm below plate A that is at zero electrical potential. It has a charge of 3,2 ×× 10-12-12 C.
    3. State whether the charge on this particle is positive or negative.
    4. Calculate the force due to the electric field on the charge.
    5. Determine the mass of the charged particle. The charge is now moved from S to Q.
    6. What is the magnitude of the force exerted by the electric field on the charge at Q?
    7. Calculate the work done when the particle is moved from S to Q.

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