Specific Charge Experiments (AQA A Level Physics): Revision Note

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 $\frac{e}{m_e}$, where $e$ is the charge of an electron and $m_e$ is the mass of an electron

• The specific charge needs to be determined in the following scenarios:

  • 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

• Recall for a circular path of radius $r$ with electrons of mass $m_e$ moving at a speed $v$, the centripetal force $F$ is:

$F=\frac{m_e{v}^2}{r}$

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

$F_B=Bev$

• This magnetic force is the centripetal force, so the two equations above can be equated and an expression for $v$ can be written:

$Be\cancel{v}=\frac{m_e{v}^{\cancel{2}}}{r}$

$v=\frac{Ber}{m_e}$

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

$e{V}_A=\frac{1}{2}{m}_e{v}^2$

• By substituting the expression for speed from above, the specific charge can be determined:

$e{V}_A=\frac{1}{2}{m}_e{\left(\frac{Ber}{m_e}\right)}^2=\frac{B^2{e}^2{r}^2}{2m_e}$

$\frac{e}{m_e}=\frac{2V_A}{B^2{r}^2}$

• The terms $V_A$, $r$, 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

• In this diagram, the Helmholtz coils' magnetic field points into the plane of the page, exerting a downward magnetic force $F_B$

  • This can be verified 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, $F_E$, is equal to $F_B$:

$F_E=F_B$

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

$F_E=eE=\frac{eV}{d}$

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

$F_B=Bev$

• By equating the two forces, the electron speed can be determined:

$B\cancel{e}v=\frac{\cancel{e}V}{d}$

$v=\frac{V}{Bd}$

• 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:

$v=\frac{Ber}{m_e}$

• Combining these two equations gave Thomson the expression:

$\frac{e}{m_e}=\frac{V}{r{B}^{2}d}$

• The terms $V$, $r$, $B$ and $d$ 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

• 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 $w$:

$t=\frac{w}{v}$

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

$a=\frac{F_E}{m_e}=\frac{eV}{d{m}_e}$

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

$s=ut+\frac{1}{2}a{t}^2$

• Initial velocity is 0 m s⁻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:

$y=\frac{1}{2}a\frac{w^2}{v^2}$

• 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 the expression for $a$ is rerranged:

$a=\frac{eV}{d{m}_e}$

• To give:

$\frac{e}{m_e}=\frac{ad}{V}$