Electrons

(a) Path in a perpendicular electric field

The electron follows a parabolic path, like a projectile. Its velocity component along its original direction stays constant. Perpendicular to it, a constant force $eE$ gives a constant acceleration $\dfrac{eE}{m}$ towards the positive plate. So $y = \dfrac{eE}{2mv^2}x^2$, which is a parabola.

(b) Undeviated electron: does it mean there is no field?

No. An undeviated electron does not prove that there is no field. It passes straight through in any of these cases:

1. Only a magnetic field, parallel to the velocity: $F = evB\sin\theta = 0$ when $\theta = 0^\circ$ or $180^\circ$.
2. Crossed electric and magnetic fields that balance: When $E$ and $B$ are perpendicular to each other and to $v$, and $eE = evB$ (that is, $v = E/B$), the two forces cancel. This is the principle of the velocity selector in Thomson's experiment.
3. Only an electric field, along the line of motion: The electron speeds up or slows down, but it is not deflected sideways.

So an undeviated path only shows that there is no net sideways force on the electron. It does not show that no field is present.

(c) Condition for no force in a magnetic field

Yes. The magnetic force is $F = evB\sin\theta$. It is zero when:

• the electron moves parallel or antiparallel to the magnetic field ($\theta = 0^\circ$ or $180^\circ$), or
• the electron is at rest ($v = 0$). A magnetic field exerts no force on a stationary charge.

i) Why the beam is undeflected

The electric and magnetic fields are crossed (perpendicular to each other and to the beam). They exert forces on the electron in opposite directions. When the two forces are equal in magnitude, $eE = evB$, the net force is zero, so the beam goes straight through undeflected.

ii) Velocity of the electrons

Given: $V = 50$ V, $d = 3$ cm $= 0.03$ m, $B = 10^{-4}$ T

1. Electric field between the plates:

$$E = \frac{V}{d} = \frac{50}{0.03} = 1.67\times10^{3}\ \text{V m}^{-1}$$

2. For no deflection, $eE = evB$, so

$$v = \frac{E}{B} = \frac{1.67\times10^{3}}{10^{-4}}$$ $$\boxed{v \approx 1.67\times10^{7}\ \text{m s}^{-1}}$$

a) Quantization of charge

Electric charge exists only in whole-number multiples of a smallest basic charge, the charge of an electron, $e = 1.6\times10^{-19}$ C. Any charge is given by

$$q = \pm ne,\qquad n = 1, 2, 3, \ldots$$

A body can have $e$, $2e$, $3e$, and so on, but never $1.5e$ or $2.7e$. The reason is that charge is carried by particles (electrons and protons), and a body can only gain or lose whole electrons. Millikan's oil drop experiment confirmed this: the charges on all the drops were integral multiples of $1.6\times10^{-19}$ C.

b) Radius of the ion's orbit

Given: $\dfrac{q}{m} = 4.40\times10^{7}$ C kg$^{-1}$, $B = 0.4$ T, $v = 3.52\times10^{5}$ m s$^{-1}$

The magnetic force provides the centripetal force:

$$qvB = \frac{mv^2}{r} \;\Rightarrow\; r = \frac{v}{(q/m)B}$$ $$r = \frac{3.52\times10^{5}}{4.40\times10^{7}\times0.4} = \frac{3.52\times10^{5}}{1.76\times10^{7}}$$ $$\boxed{r = 2.0\times10^{-2}\ \text{m} = 2\ \text{cm}}$$

c) Use of X-rays in Millikan's experiment

X-rays ionize the air between the plates. The oil drops pick up these ions (or electrons), so the charge on a drop changes. This lets the experimenter measure many different charges on the same drop, and they are all found to be multiples of $e$.

a) Terminal velocity of the oil drop

Given: $m = 3\times10^{-11}$ g $= 3\times10^{-14}$ kg, $r = 2\times10^{-4}$ cm $= 2\times10^{-6}$ m, $n = 10$, so $q = 10e = 1.6\times10^{-18}$ C, $\eta = 1.8\times10^{-5}$ N s m$^{-2}$, $g = 9.8$ m s$^{-2}$ (buoyancy of air neglected)

Viscous force (Stokes' law): $F_v = 6\pi\eta rv$, where

$$6\pi\eta r = 6\pi\times1.8\times10^{-5}\times2\times10^{-6} = 6.79\times10^{-10}\ \text{N s m}^{-1}$$

Weight: $mg = 3\times10^{-14}\times9.8 = 2.94\times10^{-13}$ N

i) No electric field

At terminal velocity the viscous force balances the weight:

$$6\pi\eta rv = mg \;\Rightarrow\; v = \frac{mg}{6\pi\eta r} = \frac{2.94\times10^{-13}}{6.79\times10^{-10}}$$ $$\boxed{v \approx 4.3\times10^{-4}\ \text{m s}^{-1}\ \text{(downward)}}$$

ii) Electric field $E = 3\times10^{5}$ V m$^{-1}$ directed downward

The drop carries excess electrons, so its charge is negative. The electric force on it is opposite to $E$, that is upward:

$$qE = 1.6\times10^{-18}\times3\times10^{5} = 4.8\times10^{-13}\ \text{N}$$

This is greater than the weight ($2.94\times10^{-13}$ N), so the drop actually moves upward. At terminal velocity the viscous force acts downward:

$$qE = mg + 6\pi\eta rv \;\Rightarrow\; v = \frac{qE - mg}{6\pi\eta r} = \frac{4.8\times10^{-13} - 2.94\times10^{-13}}{6.79\times10^{-10}}$$ $$\boxed{v \approx 2.7\times10^{-4}\ \text{m s}^{-1}\ \text{(upward)}}$$

b) Why E is kept perpendicular to B in Thomson's method

For an electron moving with velocity $v$, the magnetic force $evB$ is perpendicular to both $\vec v$ and $\vec B$. If $\vec E$ is perpendicular to both $\vec v$ and $\vec B$, then the electric force $eE$ acts along the same line as the magnetic force. With suitable polarity it acts in the opposite direction. Only then can the two forces cancel exactly, $eE = evB$, leaving the beam undeflected. This gives the speed of the electrons, $v = \dfrac{E}{B}$, which is needed to find $\dfrac{e}{m}$. If the fields were not crossed, the forces could not balance.

c) Use of X-rays in Millikan's experiment

X-rays ionize the air between the plates. The oil drops capture these ions or electrons, which changes their charge. This lets the experimenter measure several different charges on the same drop and show that each is a whole-number multiple of $e$.

Case 1: Field absent (drop falling with terminal velocity $v_1$)

          ^  F_v = 6 pi eta r v1   (viscous drag, upward)
          |
          ^  U = (4/3) pi r^3 sigma g   (buoyancy, upward)
          |
          O   oil drop (moving down, v1)
          |
          v  W = mg = (4/3) pi r^3 rho g   (weight, downward)

Balance: $mg = 6\pi\eta rv_1 + U$ (often $U$ is neglected, giving $mg = 6\pi\eta rv_1$). This gives the radius $r$ of the drop.

Case 2: Field applied (drop rising with terminal velocity $v_2$)

          ^  F_E = qE = qV/d   (electric force, upward)
          |
          ^  U   (buoyancy, upward)
          |
          O   oil drop (moving up, v2)
          |
          v  W = mg   (weight, downward)
          |
          v  F_v = 6 pi eta r v2   (viscous drag, downward)

Balance: $qE + U = mg + 6\pi\eta rv_2$. If the drop is instead held stationary, there is no viscous force and $qE = mg$ (neglecting $U$).

Conclusion of the experiment

• The charge on every oil drop was always a whole-number multiple of a smallest charge, $q = ne$. This proves that charge is quantized.
• The smallest charge, the charge of an electron, is $e = 1.6\times10^{-19}$ C.

(i) Motion of the electron and its parabolic path

The electron enters with horizontal velocity $v$ into the uniform electric field $E = V/d$ between the plates. The field points from the positive to the negative plate, so the force on the electron, $F = eE$, is towards the positive (upper) plate and perpendicular to $v$. There is no force along the horizontal direction.

Take the entry point as the origin, $x$ horizontal and $y$ towards the positive plate.

• Horizontal: uniform velocity, $x = vt$.
• Vertical: uniform acceleration $a = \dfrac{eE}{m}$ from rest, $y = \tfrac12 a t^2$.

Eliminating $t = x/v$:

$$y = \frac{eE}{2mv^2}\,x^2$$

Since $y \propto x^2$, the path is a parabola. After leaving the plates the electron moves in a straight line.

(ii) Cross field

A cross field is a region where an electric field and a magnetic field act perpendicular to each other (and to the velocity of the charge). A use is the velocity selector: charges pass undeflected only if $eE = Bev$, i.e. $v = E/B$, so only particles of this speed emerge from the slit. It is also used in Thomson's method for $e/m$.

(b) $2mg/q$

While the drop falls at terminal velocity $v$, the upward viscous drag balances the weight: $F_v = mg$ (buoyancy neglected).

To move upward at the same speed $v$, the drag now acts downward with the same magnitude $mg$, so the electric force must balance the weight plus the drag:

$qE = mg + mg \;\Rightarrow\; E = \frac{2mg}{q}$

(a) Cathode rays are always streams of electrons, whatever gas or electrode material is used, so $e/m$ has the same value ($1.76\times10^{11}$ C/kg). Positive rays are positive ions formed from the atoms or molecules of the gas in the tube. Different gases give ions of different mass, and the ions may carry different charges (one electron lost, two lost, and so on). So $e/m$ depends on the gas and differs from one ion to another.

(b) The magnetic force provides the centripetal force: $evB = mv^2/r$, so

$v = \frac{eBr}{m} = \frac{1.6\times10^{-19}\times2\times10^{-3}\times0.20}{9.11\times10^{-31}} \approx 7.0\times10^{7}\ \text{m/s}$

The potential through which it must be accelerated follows from $eV = \tfrac12mv^2$:

$V = \frac{mv^2}{2e} = \frac{9.11\times10^{-31}\times(7.03\times10^{7})^2}{2\times1.6\times10^{-19}} \approx 1.4\times10^{4}\ \text{V}$

(a) Charge on a drop moving down with constant velocity $v$

Plate A is at high positive potential and plate B is earthed, so the field $E = V/d$ points downward. The drop is negatively charged, so the electric force $qE$ on it acts upward. The drop moves down at constant speed, so the viscous drag $6\pi\eta rv$ also acts upward.

   A  + + + + + + + + + + + +   (field E points down)
            ^  qE  (electric force, upward)
            |
            ^  6 pi eta r v  (viscous drag, upward)
            |
          ( O )  drop, moving down with constant v
            |
            v  mg = (4/3) pi r^3 rho g  (weight)
   B  ------------------------  (earthed)

Since the velocity is constant, the net force is zero:

$$mg = qE + 6\pi\eta rv$$ $$\boxed{q = \frac{\tfrac{4}{3}\pi r^3\rho g - 6\pi\eta rv}{E} = \frac{d}{V}\left(\tfrac{4}{3}\pi r^3\rho g - 6\pi\eta rv\right)}$$

Here $\rho$ is the density of the oil and $\eta$ is the viscosity of air. If the buoyancy of air (density $\sigma$) is included, replace $\rho$ by $(\rho - \sigma)$.

(b) When the electric force is greater than the weight

The drop then moves upward with terminal velocity $v$, so the viscous force acts downward:

$$qE = mg + 6\pi\eta rv$$ $$\boxed{q = \frac{\tfrac{4}{3}\pi r^3\rho g + 6\pi\eta rv}{E}}$$

(c) Field needed to balance the drop

Given: $m = 1.3\times10^{-14}$ kg, $n = 4$, so $q = 4e = 4\times1.6\times10^{-19} = 6.4\times10^{-19}$ C, $g = 9.8$ m s$^{-2}$

When balanced, the drop is at rest and $qE = mg$:

$$E = \frac{mg}{q} = \frac{1.3\times10^{-14}\times9.8}{6.4\times10^{-19}}$$ $$\boxed{E \approx 2.0\times10^{5}\ \text{V m}^{-1}}$$

(a) Quantization of charge

Charge cannot take any value we like. It always occurs as a whole-number multiple of the elementary charge $e = 1.6\times10^{-19}$ C:

$$q = \pm ne,\qquad n = 1, 2, 3, \ldots$$

This is because charge is carried by electrons and protons, and a body can gain or lose only whole electrons. Charges such as $0.5e$ or $2.3e$ are never found on a free body. Millikan's oil drop experiment proved this.

(b) (i) Condition for the drop to be stationary

The upward electric force must exactly balance the downward weight of the drop (the drop is at rest, so there is no viscous force):

$$qE = mg,\qquad E = \frac{V}{d}$$

   + + + + + + + + + +   upper plate
           ^ qE = qV/d
           |
         ( O )   drop at rest
           |
           v mg
   - - - - - - - - - -   lower plate

(The drop must be negatively charged if the upper plate is positive.)

(b) (ii) Charge on the drop

Given: $mg = 1.5\times10^{-14}$ N, $d = 10$ mm $= 0.010$ m, $V = 470$ V

$$E = \frac{V}{d} = \frac{470}{0.010} = 4.7\times10^{4}\ \text{V m}^{-1}$$ $$q = \frac{mg}{E} = \frac{1.5\times10^{-14}}{4.7\times10^{4}}$$ $$\boxed{q \approx 3.2\times10^{-19}\ \text{C} \approx 2e}$$

So the drop carries 2 excess electrons, since $q/e = 3.2\times10^{-19}/1.6\times10^{-19} = 2$.

(b) (iii) If a stray alpha particle strikes the drop

• An alpha particle is a helium nucleus with charge $+2e$ and is a strong ionizer. When it hits the drop, it can change the charge on the drop, either by ionizing it (knocking out electrons) or by being captured. The charge changes by a whole multiple of $e$.
• The electric force $qE$ then no longer equals $mg$, so the balance is destroyed:
  • If the magnitude of the charge decreases (for example, $-2e + 2e = 0$), the electric force becomes smaller than the weight. The drop starts to fall, speeds up, and then falls with a constant terminal velocity.
  • If the magnitude of the charge increases, the electric force becomes larger than the weight. The drop moves upward with a terminal velocity.
• The impact may also give the drop a small momentary jolt.

To hold the drop stationary again, the p.d. must be adjusted to a new value $V' = \dfrac{mgd}{q'}$. This again shows that charge changes only in steps of $e$.