CIE AS & A Level Physics 9702: Topic 19: Capacitance- Unit : 19.1 Capacitors and capacitance Study Notes

Defining Capacitance

  • Capacitors are electrical devices used to store energy in electronic circuits, commonly for a backup release of energy if the power fails
  • They can be in the form of:
  • An isolated spherical conductor
  • Parallel plates
  •  Capacitors are marked with a value of their capacitance. This is defined as:

                                           The charge stored per unit potential difference

  • The greater the capacitance, the greater the energy stored in the capacitor
  •  A parallel plate capacitor is made up of two conductive metal plates connected to a voltage supply
  •  The negative terminal of the voltage supply pushes electrons onto one plate, making it negatively charged
  • The electrons are repelled from the opposite plate, making it positively charged
  •  There is commonly a dielectric in between the plates, this is to ensure charge does not freely flow between the plates
  • Exam Tip
    The ‘charge stored’ by a capacitor refers to the magnitude of the charge stored on each plate in a parallel plate capacitor or on the surface of a spherical conductor. The capacitor itself does not store charge.

Calculating Capacitance

  •  The capacitance of a capacitor is defined by the equation:

$
C=\frac{Q}{V}
$

  • Where:
    •  C = capacitance (F)
    •  $\mathrm{Q}=$ charge $(\mathrm{C})$
    • $\mathrm{V}=$ potential difference $(\mathrm{V})$
  • It is measured in the unit Farad (F)
    • In practice, $1 \mathrm{~F}$ is a very large unit
    •  Capacitance will often be quoted in the order of micro Farads ( $\mu \mathrm{F})$, nanofarads $(\mathrm{nF})$ or picofarads (pF)
  •  If the capacitor is made of parallel plates, $Q$ is the charge on the plates and $\mathrm{V}$ is the potential difference across the capacitor
  • The charge $\mathrm{Q}$ is not the charge of the capacitor itself, it is the charge stored on the plates or spherical conductor
    • This capacitance equation shows that an object’s capacitance is the ratio of the charge on an object to its potential

 

Capacitance of a Spherical Conductor

  •  The capacitance of a charged sphere is defined by the charge per unit potential at the surface of the sphere
  • The potential $V$ is defined by the potential of an isolated point charge (since the charge on the surface of a spherical conductor can be considered as a point charge at its centre):

$$
\mathrm{V}=\frac{Q}{4 \pi \varepsilon_0 r}
$$

  •  Substituting this into the capacitance equation means the capacitance $C$ of a sphere is given by the expression:

$$
C=4 \pi \varepsilon_0 r
$$

Worked example: Charge on parallel plates
A parallel plate capacitor has a capacitance of $1 \mathrm{nF}$ and is connected to a voltage supply of $0.3 \mathrm{kV}$. Calculate the charge on the plates.

Answer/Explanation

Step 1:

Write down the known quantities
Capacitance, $\mathrm{C}=1 \mathrm{nF}=1 \times 10^{-9} \mathrm{~F}$
Potential difference, $\mathrm{V}=0.3 \mathrm{kV}=0.3 \times 10^3 \mathrm{~V}$

Step 2:
Write out the equation for capacitance

$
\mathrm{C}=\frac{Q}{V}
$

Step 3:
Rearrange for charge Q

$
\mathbf{Q}=\mathbf{C V}
$

Step 4:
Substitute in values

$
Q=\left(1 \times 10^{-9}\right) \times\left(0.3 \times 10^3\right)=3 \times 10^{-7} \mathrm{C}=300 \mathrm{nC}
$

Exam Tip
The letter ‘ $C$ ‘ is used both as the symbol for capacitance as well as the unit of charge (coulombs). Take care not to confuse the two!

19.1.2 DERIVATION OF C = Q/V

Derivation of $\mathrm{C}=\mathrm{Q} / \mathrm{V}$

  • The circuit symbol for a parallel plate capacitor is two parallel lines
  • Capacitors can be combined in series and parallel circuits
  •  The combined capacitance depends on whether the capacitors are connected in series or parallel

Capacitors in Series

  •  Consider two parallel plate capacitors $C_1$ and $C_2$ connected in series, with a potential difference (p.d) $V$ across them
  • In a series circuit, p.d is shared between all the components in the circuit
    •  Therefore, if the capacitors store the same charge on their plates but have different p.ds, the p.d across $C_1$ is $V_1$ and across $C_2$ is $V_2$
  •  The total potential difference $V$ is the sum of $V_1$ and $V_2$

$
V=V_1+V_2
$

  •  Rearranging the capacitance equation for the p.d $V$ means $V_1$ and $V_2$ can be written as:
  • $
    \mathrm{V}_1=\frac{Q}{C_1} \quad \text { and } \quad \mathrm{V}_2=\frac{Q}{C_2}
    $
  •  Where the total p.d $V$ is defined by the total capacitance

$
\mathrm{V}=\frac{Q}{C_{\text {total }}}
$

  •  Substituting these into the equation $V=V_1+V_2$ equals:

$
\frac{Q}{C_{\text {total }}}=\frac{Q}{C_1}+\frac{Q}{C_2}
$

  • Since the current is the same through all components in a series circuit, the charge $Q$ is the same through each capacitor and cancels out
  •  Therefore, the equation for combined capacitance of capacitors in series is:
  • $
    \frac{1}{C_{\text {total }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots
    $

Capacitors in Parallel

  •  Consider two parallel plate capacitors $C_1$ and $C_2$ connected in parallel, each with p.d $V$
  •  Since the current is split across each junction in a parallel circuit, the charge stored on each capacitor is different
  • Therefore, the charge on capacitor $C_1$ is $Q_1$ and on $C_2$ is $Q_2$
  • The total charge $Q$ is the sum of $Q_1$ and $Q_2$

$
\mathbf{Q}=\mathbf{Q}_1+\mathbf{Q}_2
$

  • Rearranging the capacitance equation for the charge $Q$ means $Q_1$ and $Q_2$ can be written as:

$
Q_1=C_1 \mathbf{V} \text { and } Q_2=C_2 \mathbf{V}
$

  •  Where the total charge $Q$ is defined by the total capacitance:

$
\mathbf{Q}=\mathbf{C}_{\mathrm{total}} \mathbf{V}
$

  •  Substituting these into the $\mathrm{Q}=\mathrm{Q}_1+\mathrm{Q}_2$ equals:

$
C_{\text {total }} V=C_1 V+C_2 V=\left(C_1+C_2\right) V
$

  • Since the p.d is the same through all components in each branch of a parallel circuit, the p.d $V$ cancels out
  • Therefore, the equation for combined capacitance of capacitors in parallel is:

$
\mathrm{C}_{\mathrm{total}}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \ldots
$

Exam Tip
You will be expected to remember these derivations for your exam, therefore, make sure you understand each step. You should especially make sure to revise how the current and potential difference varies in a series and parallel circuit.

19.1.3 CAPACITORS IN SERIES & PARALLEL

Capacitors in Series  & Parallel

  •  Recall the formula for the combined capacitance of capacitors in series:

$
\frac{1}{C_{\text {total }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots
$

  •  In parallel:

$
\mathrm{C}_{\mathrm{total}}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \ldots
$

 

Worked example: Calculating total capacitance between two points

Three capacitors with a capacitance of $23 \mu \mathrm{F}, 35 \mu \mathrm{F}$ and $40 \mu \mathrm{F}$ are connected as shown below.

                             

Calculate the total capacitance between points $\mathrm{A}$ and $\mathrm{B}$.

Answer/Explanation

Step 1:
Calculate the combined capacitance of the two capacitors in parallel
Capacitors in parallel: $\mathrm{C}_{\text {total }}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \ldots$

$
C_{\text {parallel }}=23+35=58 \mu \mathrm{F}
$

Step 2:
Connect this combined capacitance with the final capacitor in series
Capacitors in series: $\frac{1}{C_{\text {total }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} \ldots$

$
\frac{1}{C_{\text {total }}}=\frac{1}{58}+\frac{1}{40}=\frac{49}{1160}
$

Step 3:
Rearrange for the total capacitance

$
C_{\text {total }}=\frac{1160}{49}=23.673 \ldots=24 \mu F(2 \text { s.f) }
$

Exam Tip
Both the combined capacitance equations look similar to the equations for combined resistance in series and parallel circuits. However, take note that they are the opposite way around to each other!

19.1.4 AREA UNDER A POTENTIAL-CHARGE GRAPH

Area Under a Potential-Charge Graph

  • When charging a capacitor, the power supply pushes electrons from the positive to the negative plate
    •  It therefore does work on the electrons, which increase their electric potential energy
  • At first, a small amount of charge is pushed from the positive to the negative plate, then gradually, this builds up
    • Adding more electrons to the negative plate at first is relatively easy since there is little repulsion
  •  As the charge of the negative plate increases ie. becomes more negatively charged, the force of repulsion between the electrons on the plate and the new electrons being pushed onto it increases
  • This means a greater amount of work must be done to increase the charge on the negative plate or in other words:

The potential difference $V$ across the capacitor increases as the amount of charge $Q$ increases

  •  The charge $Q$ on the capacitor is directly proportional to its potential difference $V$
  • The graph of charge against potential difference is therefore a straight line graph through the origin
  • The electric potential energy stored in the capacitor can be determined from the area under the potential-charge graph which is equal to the area of a right-angled triangle:

Area $=\frac{1}{2} \times$ base $\times$ height

 

Worked example: Potential charge

The variation of the potential $V$ of a charged isolated metal sphere with surface charge $Q$ is shown on the graph below.

                                 

Using the graph, determine the electric potential energy stored on the sphere when charged to a potential of $100 \mathrm{kV}$.

Answer/Explanation

Step 1:
Determine the charge on the sphere at the potential of 100 kV

                               

From the graph, the charge on the sphere at $100 \mathrm{kV}$ is $1.8 \boldsymbol{\mu C}$
Step 2:
Calculate the electric potential energy stored
The E.P.E stored is the area under the graph at $100 \mathrm{kV}$
The area is equal to a right-angled triangle, so, can be calculated with the equation:

$
\text { Area }=\frac{1}{2} \times \text { base } \times \text { height }
$
$
\text { Area }=\frac{1}{2} \times 1.8 \mu \mathrm{C} \times 100 \mathrm{kV}
$
$
\text { E.P.E }=\frac{1}{2} \times 1.8 \times 10^{-6} \times 100 \times 10^3=0.09 \mathrm{~J}
$

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