Home / CIE AS & A Level Physics 9702: Topic 18: Electric fields- Unit : 18.5 Electric potential Study Notes

CIE AS & A Level Physics 9702: Topic 18: Electric fields- Unit : 18.5 Electric potential Study Notes

ELECTRIC FORCE BETWEEN TWO POINT CHARGES

Coulomb’s Law

  •  All charged particles produce an electric field around it
    • This field exerts a force on any other charged particle within range
  • The electrostatic force between two charges is defined by Coulomb’s Law
    •  Recall that the charge of a uniform spherical conductor can be considered as a point charge at its centre
  •  Coulomb’s Law states that:
                            The electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation
  •  The Coulomb equation is defined as:

$
\mathrm{F}_{\mathrm{E}}=\frac{Q_1 Q_2}{4 \pi \varepsilon_0 r^2}
$

  • Where:
    • $F_E=$ electrostatic force between two charges (N)
    •  $\mathrm{Q}_1$ and $\mathrm{Q}_2=$ two point charges (C)
    •  $\varepsilon_0=$ permittivity of free space
    •  $r=$ distance between the centre of the charges $(\mathrm{m})$
  •  The $1 / r^2$ relation is called the inverse square law
    • This means that when a charge is twice as far as away from another, the electrostatic force between them reduces by $(1 / 2)^2=1 / 4$
  • If there is a positive and negative charge, then the electrostatic force is negative, this can be interpreted as an attractive force
  • If the charges are the same, the electrostatic force is positive, this can be interpreted as a repulsive force
  • Since uniformly charged spheres can be considered as point charges, Coulomb’s law can be applied to find the electrostatic force between them as long as the separation is taken from the centre of both spheres

Worked example: Coulomb’s Law
An alpha particle is situated $2.0 \mathrm{~mm}$ away from a gold nucleus in a vacuum. Assuming them to be point charges, calculate the magnitude of the electrostatic force acting on each of the charges.
Atomic number of helium $=2$
Atomic number of gold $=79$
Charge of an electron $=1.60 \times 10^{-19} \mathrm{C}$

Answer/Explanation

Step 1:
Write down the known quantities

  • Distance, $r=2.0 \mathrm{~mm}=2.0 \times 10^{-3} \mathrm{~m}$

The charge of one proton $=+1.60 \times 10^{-19} \mathrm{C}$
An alpha particle (helium nucleus) has 2 protons

  • Charge of alpha particle, $Q_1=2 \times 1.60 \times 10^{-19}=+3.2 \times 10^{-19} \mathrm{C}$

The gold nucleus has 79 protons
Charge of gold nucleus, $Q_2=79 \times 1.60 \times 10^{-19}=+1.264 \times 10^{-17} \mathrm{C}$

Step 2:

The electrostatic force between two point charges is given by Coulomb’s Law

$
\mathrm{F}_{\mathrm{E}}=\frac{Q_1 Q_2}{4 \pi \varepsilon_0 r^2}
$

Step 3:
Substitute values into Coulomb’s Law

$F_E=\frac{\left(3.2 \times 10^{-19}\right) \times\left(1.264 \times 10^{-17}\right)}{\left(4 \pi \times 8.85 \times 10^{-12}\right) \times\left(2.0 \times 10^{-3}\right)^2}=9.092 . . \times 10^{-21}=9.1 \times 10^{-21} \mathbf{N}$ (2 s.f.)

Electric Potential

  •  In order to move a positive charge closer to another positive charge, work must be done to overcome the force of repulsion between them
  •  Energy is therefore transferred to the charge that is being pushed upon
    • This means its potential energy increases
  •  If the positive charge is free to move, it will start to move away from the repelling charge
    • As a result, its potential energy decreases back to 0
  •  This is analogous to the gravitational potential energy of a mass increasing as it is being lift upwards and decreasing and it falls
  • The electric potential at a point is defined as:
  • The work done per unit positive charge in bringing a small test charge from infinity to a defined point
  •  Electric potential is a scalar quantity
  • This means it doesn’t have a direction
  •  However, you will still see the electric potential with a positive or negative sign. This is because the electric potential is:
    • Positive when near an isolated positive charge
    • Negative when near an isolated negative charges
    • Zero at infinity
  •  Positive work is done by the mass from infinity to a point around a positive charge and negative work is done around a negative charge. This means:
    •  When a test charge moves closer to a negative charge, its electric potential decreases
    • When a test charge moves closer to a positive charge, its electric potential increases
  •  To find the potential at a point caused by multiple charges, add up each potential separately

Electric Potential Due to a Point Charge

  • The electric potential in the field due to a point charge is defined as:

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

  •  Where:
    • $\mathrm{V}=$ the electric potential (V)
    •  $\mathrm{Q}=$ the point charge producing the potential (C)
    •  $\varepsilon_0=$ permittivity of free space $\left(\mathrm{F} \mathrm{m}^{-1}\right)$
    •  $r=$ distance from the centre of the point charge $(m)$
  •  This equation shows that for a positive $(+)$ charge:
    • As the distance from the charge $r$ decreases, the potential $V$ increases
    • This is because more work has to be done on a positive test charge to overcome the repulsive force
  •  For a negative (-) charge:
  • As the distance from the charge $r$ decreases, the potential $V$ decreases
  • This is because less work has to be done on a positive test charge since the attractive force will make it easier
  • Unlike the gravitational potential equation, the minus sign in the electric potential equation will be included in the charge
  • The electric potential changes according to an inverse square law with distance
  •  Note: this equation still applies to a conducting sphere. The charge on the sphere is treated as if it concentrated at a point in the sphere from the point charge approximation

Worked example: Van de Graaf charge

 A Van de Graaf generator has a spherical dome of radius $15 \mathrm{~cm}$. It is charged up to a potential of $240 \mathrm{kV}$.
Calculate:
a) The charge is stored on the dome.
b) The potential at a distance of $30 \mathrm{~cm}$ from the dome.

Part (a)
Step 1:   $\quad$ Write down the known quantities
Radius of the dome, $r=15 \mathrm{~cm}=15 \times 10^{-2} \mathrm{~m}$
Potential difference, $\mathrm{V}=240 \mathrm{kV}=240 \times 10^3 \mathrm{~V}$

Step 2:
Write down the equation for the electric potential due to a point charge

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

Step 3:
Rearrange for charge Q

$
Q=V 4 \pi \varepsilon_0 r
$

Step 4:
Substitute in values

$
Q=\left(240 \times 10^3\right) \times\left(4 \pi \times 8.85 \times 10^{-12}\right) \times\left(15 \times 10^{-2}\right)=4.0 \times 10^{-6} \mathrm{C}=4.0 \mu C
$

Part (b)
Step 1: Write down the known quantities

$
Q=\text { charge stored in the dome }=4.0 \mu \mathrm{C}=4.0 \times 10^{-6} \mathrm{C}
$
$r=$ radius of the dome + distance from the dome $=15+30=45 \mathrm{~cm}=45 \times 10^{-2} \mathrm{~m}$

Step 2:
Write down the equation for electric potential due to a point charge

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

Step 3:
Substitute in values

$
V=\frac{\left(4.0 \times 10^{-6}\right)}{\left(4 \pi \times 8.85 \times 10^{-12}\right) \times\left(45 \times 10^{-2}\right)}=79.93 \times 10^3=80 \mathrm{kV}(2 \text { s.f. })
$

Potential Gradient

  •  An electric field can be defined in terms of the variation of electric potential at different points in the field:
    The electric field at a particular point is equal to the negative gradient of a potential-distance graph at that point
  • The potential gradient is defined by the equipotential lines
    •  These demonstrate the electric potential in an electric field and are always drawn perpendicular to the field lines
  •  Equipotential lines are lines of equal electric potential
    • Around a radial field, the equipotential lines are represented by concentric circles around the charge with increasing radius
    •  The equipotential lines become further away from each other
    •  In a uniform electric field, the equipotential lines are equally spaced
  •  The potential gradient in an electric field is defined as:
    The rate of change of electric potential with respect to displacement in the direction of the field
  • The electric field strength is equivalent to this, except with a negative sign:

$
E=-\frac{\Delta V}{\Delta r}
$

  • Where:
    • $E=$ electric field strength $\left(\mathrm{V} \mathrm{m}^{-1}\right)$
    • $\Delta \mathrm{V}=$ change in potential (V)
    • $\Delta \mathrm{r}=$ displacement in the direction of the field $(\mathrm{m})$
  • The minus sign is important to obtain an attractive field around a negative charge and repulsive field around a positive charge
  •  The electric potential changes according to the charge creating the potential as the distance $r$ increases from the centre:
    • If the charge is positive, the potential decreases with distance
    •  If the charge is negative, the potential increases with distance
  •  This is because the test charge is positive

Exam Tip
One way to remember whether the electric potential increases or decreases with respect to the distance from the charge is by the direction of the electric field lines. The potential always decreases in the same direction as the field lines and vice versa.

Electric Potential Energy of Two Point Charges

  •  The electric potential energy $\mathrm{E}_{\mathrm{p}}$ at point in an electric field is defined as:
    The work done in bringing a charge from infinity to that point
  •  The electric potential energy of a pair of point charges $Q_1$ and $Q_2$ is defined by:
  • $
    \mathrm{E}_{\mathrm{p}}=\frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}
    $
  • Where:
    •  $\mathrm{E}_{\mathrm{p}}=$ electric potential energy $(\mathrm{J})$
    • $\mathrm{r}=$ separation of the charges $\mathrm{Q}_1$ and $\mathrm{Q}_2(\mathrm{~m})$
    •  $\varepsilon_0=$ permittivity of free space $\left(\mathrm{F} \mathrm{m}^{-1}\right)$
  • The potential energy equation is defined by the work done in moving point charge $Q_2$ from infinity towards a point charge $Q_1$.
  •  The work done is equal to:

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

  • Where:
    • $W=$ work done $(J)$
    • $\mathrm{V}=$ electric potential due to a point charge $(\mathrm{V})$
    • $\mathrm{Q}=$ Charge producing the potential (C)
  • This equation is relevant to calculate the work done due on a charge in a uniform field
  • Unlike the electric potential, the potential energy will always be positive
  • Recall that at infinity, $\mathrm{V}=0$ therefore $\mathrm{E}_{\mathrm{p}}=0$
  • It is more useful to find the change in potential energy eg. as one charge moves away from another
  • The change in potential energy from a charge $Q_1$ at a distance $r_1$ from the centre of charge $Q_2$ to a distance $r_2$ is equal to:

$\Delta \mathrm{E}_{\mathrm{p}}=\frac{Q_1 Q_2}{4 \pi \varepsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

  • The change in electric potential $\Delta V$ is the same, without the charge $Q_2$

$
\Delta \mathrm{V}=\frac{Q}{4 \pi \varepsilon_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)
$

  •  Both equations are very similar to the change in gravitational potential between two points near a point mass

Worked example: Alpha particle

An a-particle ${ }_2^4 \mathrm{He}$ is moving directly towards a stationary gold nucleus ${ }_{79}^{197} \mathrm{Au}$. At a distance of $4.7 \times 10^{-15} \mathrm{~m}$, the a-particle momentarily comes to rest. Calculate the electric potential energy of the particles at this instant.

Answer/Explanation

Step 1:

Write down the known quantities

  • Distance, $r=4.7 \times 10^{-15} \mathrm{~m}$

The charge of one proton $=+1.60 \times 10^{-19} \mathrm{C}$
An alpha particle (helium nucleus) has 2 protons

  • Charge of alpha particle, $Q_1=2 \times 1.60 \times 10^{-19}=+3.2 \times 10^{-19} \mathrm{C}$
    The gold nucleus has 79 protons
  • Charge of gold nucleus, $Q_2=79 \times 1.60 \times 10^{-19}=+1.264 \times 10^{-17} \mathrm{C}$

Step 2:
Write down the equation for electric potential energy

$
\mathrm{E}_{\mathrm{p}}=\frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}
$

Step 3:
Substitute values into the equation

$
E_p=\frac{\left(1.264 \times 10^{-17}\right) \times\left(3.2 \times 10^{-19}\right)}{\left(4 \pi \times 8.85 \times 10^{-12}\right) \times\left(4.7 \times 10^{-15}\right)}=7.7 \times 10^{-12} \mathrm{~J}(2 \mathrm{~s} . \mathrm{f})
$

Exam Tip
When calculating electric potential energy, make sure you do not square the distance!

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