Specific Charge Experiments (AQA A Level Physics)

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Dan MG

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Dan MG

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Determining the Specific Charge of an Electron

  • Specific charge is the charge per unit mass of an object - this has units of C kg-1
    • For an electron, this is e over m subscript e, where  is the charge of an electron and m is the mass of an electron
  • You need to be able to describe how specific charge is determined using:
    • A magnetic field only
    • A magnetic field and an electric field
    • An electric field only

Determining Specific Charge with A Magnetic Field Only

  • When moving in a magnetic field, a charge experiences a force perpendicular to its motion

Electron beam in a magnetic field

12-1-3-circular-path

The radius of the electron beam's circular path can be measured to calculate specific charge

  • Recall for a circular path of radius with electrons of mass mmoving at a speed v, the centripetal force  is:

F space equals space fraction numerator m subscript e v squared over denominator r end fraction

  • For a magnetic force, FB, from a field with flux density  on a charge of magnitude e :

F subscript B space equals space B e v

  • This magnetic force is the centripetal force so we can equate the two equations above and write as an expression for v :

B e up diagonal strike v space equals space fraction numerator m subscript e v to the power of up diagonal strike 2 end exponent over denominator r end fraction

v space equals space fraction numerator B e r over denominator m subscript e end fraction

  • Recall from Thermionic Emission that work done on electrons by an anode of potential difference VA can be equated to their kinetic energy:

e V subscript A space equals space 1 half m subscript e v squared

  • By substituting the expression for speed from above, we can determine the specific charge (try deriving this yourself):

e over m subscript e space equals space fraction numerator 2 V subscript A over denominator B squared r squared end fraction

  • The terms VAr, and B  can all be measured, so this experiment allows the specific charge of the electron to be determined

Balancing Electric and Magnetic Fields - J.J. Thomson's Method

  • To determine the specific charge of an electron, J.J. Thomson balanced magnetic and electric forces on a beam of electrons
    • A piece of apparatus called Helmholtz coils generates a magnetic field
    • Oppositely charged plates generate an electric field

A diagram showing the balanced forces from Helmholtz coils and an electric field

12-1-3-helmholtz

The magnetic field generates a downward force, the electric field between the plates generates an upward force - the fluorescent screen shows a straight beam when these two forces are equal and opposite

  • In this diagram, the Helmholtz coils' magnetic field points into the plane of the page, exerting a downward magnetic force F (you can verify this with the left-hand rule)
  • The electric field strength, E, can be varied by changing the potential difference V  across the plates (separated by distance d ) until the electric force, FE, is equal to FB :

F subscript E space equals space F subscript B

  • Recall that the force from the electric field can be calculated using the charge on the electron, e :

F subscript E space equals space e E space equals space fraction numerator e V over denominator d end fraction

  • For magnetic flux density  and electron velocity v, magnetic force is:

F subscript B space equals space B e v

  • Equating the two forces allows us to determine electron speed:

B up diagonal strike e v space equals space fraction numerator up diagonal strike e V over denominator d end fraction

v space equals space fraction numerator V over denominator B d end fraction

  • The electric field was then switched off, and the beam formed a circular path, allowing the equation from the magnetic field only method to be used:

v space equals space fraction numerator B e r over denominator m subscript e end fraction

  • Combining these two equations gave Thomson the expression:

e over m subscript e space equals space fraction numerator V over denominator r B squared d end fraction

  • The terms V r B  and can all be measured, so this experiment allowed the specific charge of the electron to be determined

Specific Charge using an Electric Field Only

  • This method uses constant acceleration equations for the beam of electrons passing through an electric field

Trajectory of electron beam in electric field

12-1-3-electric-field

The horizontal speed of the electrons remain constant, but they have a constant acceleration vertically

  • This can be treated like a standard trajectory problem - time t  of the path is calculated using the horizontal speed v  and width of the plates :

t space equals space w over v

  • Now considering the vertical plane, the upwards acceleration a  can be calculated using F = ma :

a space equals space F subscript E over m subscript e space equals space fraction numerator e V over denominator d m subscript e end fraction

  • Constant acceleration equations can also be used in the vertical plane for another expression for acceleration:

s space equals space u t space plus space 1 half a t squared

    • Initial velocity is 0 ms−1 vertically, therefore ut  = 0
    • Displacement, s, can be measured (see y  on the diagram)
    • Time,t, is given by the first expression above
    • Therefore:

y space equals space 1 half a w squared over v squared

  • Speed can be calculated by equating work done and kinetic energy, as in the magnetic field method, and acceleration can be calculated once y and w are measured
    • This means specific charge can once again be determined if we rearrange this expression for a :

a space equals space fraction numerator e V over denominator d m subscript e end fraction

    • To give:

e over m subscript e space equals space fraction numerator a d over denominator V end fraction

Worked example

A group of students are unsure of the magnetic flux density generated by an old set of Helmholtz coils from their school.

Using a pair of charged plates of known separation with a known potential difference across them and the Helmholtz coils, they form a straight beam coming from an electron gun. They do not know the potential difference across the electron gun.

Their teacher reminds them of the following equations:

specific space charge space of space an space electron space equals space fraction numerator v over denominator r B end fraction

v space equals space fraction numerator V over denominator B d end fraction

Their teacher also reminds them of their AQA data booklets.

Explain what else they must do to determine the magnetic flux density of the Helmholtz coils.

Answer:

Step 1: List the quantities the students know

  • The students know:
    • Potential difference, V
    • Separation of the plates, d
  • The students need to find the term B

Step 2: Describe how the students can determine the missing quantities

  • The students need the radius r  of the electron beam in the electric field
  • They can turn off the electric field and measure the radius of the circular path of the electrons
  • Specific charge of an electron is the magnitude of an electron's charge divided by its mass
  • Both quantities can be found in their data booklets

Step 3: Explain how they can use this to determine B

  • The students must now rearrange the given equations to make the subject and substitute their values

Examiner Tip

You must be prepared to answer questions and perform calculations on all three methods, but you only need to be able to recall the method for one of the three methods. The magnetic field method has appeared regularly in past papers so this may be a sensible choice.

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Dan MG

Author: Dan MG

Expertise: Physics

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to SME. Here, he carries on his passion for writing enjoyable physics questions and helping young people to love physics.