Capacitance & Dielectrics

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This is a topic from Higher Physics 1B


Defining Capacitors

A capacitor is a type of device used to store energy in the form of electric charge. Capacitors consist of two conductors, called 'plates', such that when a potential difference is applied across the plates, by a battery for example, positive charge builds up on one plate and negative charge of equal magnitude builds up on the other.

Defining Capacitance

For a given capacitor, the capacitance is defined as the ratio of the magnitude of the charge stored on either plate to the potential difference between the plates:

C = Q / ΔV

Where C represents capacitance in farads (F)

Q represents charge in Coulombs (C)
ΔV represents the electric potential difference in Volts (V)
  • By dimensional analysis one volt equates to one joule-per-coulomb (JC-1). Thus because one farad equates to one coulomb-per-volt (CV-1), it can be expressed as being equivalent to one coulomb-squared-per-joule (C2J-1)
  • Capacitance is a positive quantity
  • Typical capacitances are closer to the order of magnitude of microfarads (μF) than farads.

Calculating Capacitance

The general method in deriving an expression for the the capacitance of charged objects is to

  1. Find the electric field
  2. Find the potential difference
  3. Calculate the capacitance

Summary of Results

Screen Shot 2012-08-06 at 2.43.42 PM.png

Parallel Plate Capacitor

Conceptualise the Problem

Consider the setup in the following diagram: Screen Shot 2012-08-06 at 12.58.59 PM.png

  • The potential difference between the terminals of the battery establishes an electric field in the wires
  • The field applies a force on the electrons in the wire such that they flow away from the negative terminal towards the plate on the right, and away from the plate on the left towards the positive terminal
  • This continues until there is equilibrium, and thus no electric field in the wire, and no movement of electrons
  • The two plates are now charged with the same potential difference as between the two terminals of the battery

Solve Quantitatively

  • The charge density on the surface of the plates is given by:
σ = Q / A
  • Meanwhile, by Gauss's Law it can be shown that E = σ / εo and so
E = Q / Aεo finding the electric field
  • It was shown in the Electric Potential page that V = Ed for two oppositely charges plates, and so
ΔV = Qd / Aεo finding the voltage
  • Substituting into the definition of capacitance gives
C = Q / ΔV = QAεo / Qd finding the capacitance
= Aεo / d

Cylindrical Capacitor

Consider the setup in the following diagram: Screen Shot 2012-08-06 at 1.44.22 PM.png

  • The electric field between the cylinders can be calculated using Gauss's law by imagining the Gaussian surface of a concentric cylinder with radius r between the two plates:
E.dA = q / εo
E dA = q / εo
E(2πrL) = q / εo (as the circular surfaces of the cylinder are parallel to field lines while the curved surface is perpendicular)
q = E(2πrL)εo
  • The charge density along the length of the cylinders is given by λ = dq / dL
dq / dL = E(2πr)εo
λ = E(2πr)εo
E = λ / 2πrεo
  • Voltage can then be calculated by the definition

Screen Shot 2012-07-30 at 11.16.07 PM.png as follows:
Screen Shot 2012-08-06 at 2.09.27 PM.png

  • Thus capacitance can be calculated:
C = Q / ΔV
= Q / (2keλ)(ln(b / a))
= L / 2keln(b / a) (by the definition of λ)

Isolated Spherical Charge

Counter-intuitively an isolated spherical charge has a capacitance if an imaginary spherical shell of infinite radius and voltage of zero is taken as the second 'plate' in the capacitor componen.

  • The electric potential of a sphere of radius a is given by keQ / a and so the potential difference between the plates is given by
ΔV = VA - VB
= keQ / a - 0
= keQ / a
  • And so the capacitance is given by
C = Q / ΔV
= a / ke
= 4πεoa

Spherical Capacitor

  • For two concentric spheres of radii a and b, and charges +Q and -Q respectively, the electric field is given simply by the inverse square rule with respect to the inner sphere:
E = keQ / r2
  • Thus potential difference can be found as follows:

Screen Shot 2012-08-06 at 2.35.29 PM.png

  • This gives us the capacitance:

Screen Shot 2012-08-06 at 2.35.37 PM.png
This result confirms our previous derivation for an isolated sphere, because as b tends towards infinity the capacitance tends towards 4πεoa

Capacitors in Series and Parallel

  • When multiple capacitors are placed in a circuit in parallel the total capacitance for the circuit is equal to the sum of the individual capacitances:
CTotal = C1 + C2 + ... + Cn
  • When multiple capacitors are placed in a circuit in series the inverse of the total capacitance is equal to the sum of the inverses of the individual capacitances:
1 / CTotal = 1 / C1 + 1 / C2 + ... + 1 / Cn
  • The circuit diagram symbols for capacitors and other elements are seen in the figure below:

Screen Shot 2012-08-06 at 2.47.41 PM.png

Derivation for Capacitance in Parallel

Consider the following setup: Screen Shot 2012-08-06 at 3.01.20 PM.png

  • The total charge in the plates equals the sum of the individual charges:
QTotal = Q1 + Q2
  • The potential difference across each capacitor is the same as that over the battery:
ΔV = V1 = V2
  • By the definition of capacitance:
CTotal = QTotal / ΔV
= (Q1 + Q2) / ΔV
= Q1 / ΔV + Q2 / ΔV
= Q1 / V1 + Q2 / V2
= C1 + C2

Derivation for Capacitance in Series

Consider the following setup: Screen Shot 2012-08-06 at 10.41.34 PM.png

  • The left plate of C1 and the right plate of C2 are charged by the potential difference of the battery, and the electric field created by these two charged plates charges the remaining plates such that all the plates have a charge of the same magnitude:
Q = Q1 = Q2
  • The potential differences across each capacitor are not necessarily equal, but their sum clearly must equal the total potential difference in the circuit:
ΔVTotal = V1 + V2
  • By the definition of capacitance:
ΔVTotal = Q / CTotal
  • Combining this with the previous expression for voltage gives:
Q / CTotal = Q1 / C1 + Q2 / C2
= Q / C1 + Q / C2
1 / CTotal = 1 / C1 + 1 / C2

Capacitors and Potential Energy

As stated in their definition, capacitors store electrical potential energy which arises from the separation of opposite charges which are attracted to one another. This electric potential energy comes from the chemical potential energy in the battery, prompting the question of how much work needs to be done by the battery to charge the capacitor's plates to the potential difference of the battery. The result in its various forms is as follows:

W = U = Q2 / 2C = QΔV / 2 = C(ΔV)2 / 2

This result is derived using the definitions of voltage and capacitance:

  • The relationship between work and potential energy is as follows:
dW = dU
  • Using the definition of voltage:
dW = ΔV dq
  • Using the definition of capacitance:
dW = (q / C) dq
  • Integrating both sides gives the result:
Screen Shot 2012-08-06 at 11.16.09 PM.png

Substituting the derived capacitance and electric field between two charged plates (ΔV = Ed and C = εoA/d respectively) allows the result to be rearranged as follows:

U = εoAdE2 / 2

Because the volume occupied by the electric field is given by Ad, the energy per unit volume, uE = U / V is given by:

uE = εoE2 / 2

This shows that the energy density at any point in an electric field is proportional to the square of the magnitude of the electric field at that point.


A dielectric is an insulator which when placed between the plates of a capacitor increases its capacitance.

  • Dielectrics have the following advantages:
    • Increased capacitance
    • Increased maximum operating voltage
    • Maintain the physical separation between the plates which are attracted to one another
  • Dielectric strength refers to the magnitude of the voltage that can be applied to the capacitor without discharge occurring through the insulator
  • Dielectrics increase the capacitance between two charged plates by a Dielectric Constant κ which is dependant upon the specific material which makes up the dielectric
  • The theory behind the functionality of dielectrics is as follows:
    • Dielectrics increase capacitance by decreasing the electric field - the smaller the electric field the smaller the electrical potential energy between the plates, the smaller the potential difference between the plates, the greater the capacitance
    • Dielectrics decrease the electric field because the external electric field exerts a force on the molecules within the dielectric such that they rotate to be aligned with the electric field, inducing a surface charge on the insulator which causes an induced electric field which acts in opposition to the external field. This is seen in the following diagram:
  • Thus dielectrics increase the maximum operating voltage but decrease the potential energy stored

Screen Shot 2012-08-06 at 11.35.55 PM.png

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