18.2 Uniform Electric Field

Field Strength between Parallel Plates

Derivation

A potential diffrerence, \Delta \mathrm{V} applied across two metal plates.

When a potential difference, $\Delta \mathrm{V}$, is applied across two metal plates, a uniform electric field is formed between them:

• The field lines are equally spaced and parallel.
• A point charge placed anywhere inside this uniform field experiences the same electric field strength and force.

Consider a positive test charge with charge $q$ placed just next to the positive plate:

• The charge $q$ experiences a force $\mathrm{F}$ that accelerates it towards the negative plate.

• The separation of the two plates is $\Delta \mathrm{d}$.

• The work done, $\mathrm{W}$, by the force $\mathrm{F}$ in accelerating the charge $q$ to the negative plate is given by: $\mathrm{W}=\mathrm{F\Delta d}$
  

  

  The work done in moving a charge from one plate to another depends on the potential difference and distance.

• The potential difference, $\mathrm{\Delta V}$, across the plates is defined as the work done per unit positive charge in moving a charge $q$ from one plate to another: $\mathrm{\Delta V}=\frac{\mathrm{W}}{q}$
• This relationship can be rearrange to make the work done subject of formula: $\mathrm{W}=q\mathrm{\Delta V}$
• We have $\mathrm{W}=\mathrm{F\Delta d}$, $\mathrm{W}=q\mathrm{\Delta V}$ and from the definition of electric field strength: $\mathrm{F}=q\mathrm{E}$. Putting these together: $\begin{array}{rl}\mathrm{F\Delta }\mathrm{d}& =q\mathrm{\Delta V}\\ \mathrm{E}q\mathrm{\Delta d}& =q\mathrm{\Delta V}\\ \mathrm{E}\mathrm{\Delta d}& =\mathrm{\Delta V}\\ \mathrm{E}& =\frac{\mathrm{\Delta V}}{\mathrm{\Delta d}}\end{array}$

Therefore, the electric field strength of a uniform electric field between two plates is given by: $\boxed{\mathrm{E}=\frac{\mathrm{\Delta V}}{\mathrm{\Delta d}}}$ Units: $\mathrm{V}{\mathrm{m}}^{-1}$ or $\mathrm{N}{\mathrm{C}}^{-1}$

Motion of Charged Particles in a Uniform Field

Consider a horizontal beam of electrons (negatively charged) travelling at constant speed into a uniform electric field created between two horizontal parallel metal plates in a vacuum:

The path of electrons entering a uniform electric field.

Critical

The path of the electrons outside the electric field (before entering and after exiting the plates) is a straight line.

• The electron continues through the field with a constant horizontal component of velocity, $v_x$.
• The electron’s vertical component of speed, $v_y$, increases due to the acceleration produced by the upwards force $\mathrm{F}$ exerted on the electron by the electric field.
• The force $\mathrm{F}$ experienced by a charged particle in the electric field is given by:

$$\mathrm{F}=\mathrm{E}\mathrm{q}$$

• From Newton’s Second Law:

$$\begin{array}{rl}\mathrm{F}& =ma\\ a& =\frac{\mathrm{F}}{m}\\ a& =\frac{\mathrm{E}\mathrm{q}}{m}\end{array}$$

• The time, $t$, for a particle to travel the length, $x$, of the plates is given by:

$$t=\frac{\text{distance}}{\text{horizontal velocity}}=\frac{x}{v_x}$$

• Since the acceleration is constant, the equation $s=ut+\frac{1}{2}a{t}^2$ can be used to determine the vertical displacement, $y$.
• Since the initial vertical velocity of the particle is zero, the equation can be rewritten as $s=\frac{1}{2}a{t}^2$:

$$\begin{array}{rl}s& =\frac{1}{2}a{t}^2\\ y& =\frac{1}{2}\times \frac{\mathrm{E}\mathrm{q}}{m}\times {\left(\frac{x}{v_x}\right)}^2\end{array}$$

• The equation $v=u+at$ can be used to determine the vertical velocity, $v_y$, of the particle as it leaves the electric field, since initial vertical velocity is zero, the equation can be rewritten as $v=at$:

$$\begin{array}{rl}v& =at\\ v_y& =\frac{\mathrm{E}\mathrm{q}}{m}\times \frac{x}{v_x}\end{array}$$

Worked Example

9702_s07_qp_2, Question 2

Determine whether the electron will emerge from between the two plates or will hit the top plate.

Logic: We need to determine the vertical displacement, y, of the electron as it leaves the plates:

• If the vertical displacement is greater than the distance from the centre line to the top plate, the electron hit the top plate.
• If the vertical displacement is smaller than the distance from the centre line to the top plate, the electron emerges from between the plates.

Step 1: Calculate the time, $t$, for the electron to travel the length of the plates $\begin{array}{rl}t& =\frac{\text{length of plate}}{v_x}\\ t& =\frac{0.120}{5.0\times 10^7}\\ & =2.4\times 10^{-9}s\end{array}$ Step 2: Calculate the vertical acceleration $\begin{array}{rl}\mathrm{F}& =ma\\ a& =\frac{\mathrm{F}}{m}\\ \mathrm{F}& =\mathrm{E}\mathrm{q}\\ a& =\frac{\mathrm{E}\mathrm{q}}{m}\\ \mathrm{E}& =\frac{\mathrm{\Delta V}}{\Delta d}\\ a& =\frac{\left(\frac{\mathrm{\Delta V}}{\Delta d}\right)\times e}{m_e}\\ & =\frac{\left(\frac{210}{0.015}\right)\times 1.6\times 10^{\text{-19}}}{9.11\times 10^{-31}}\\ & =2.459\times 10^{15}\mathrm{m}{\mathrm{s}}^{-2}\end{array}$ Step 3: Calculate the vertical displacement $\begin{array}{rl}s& =\frac{1}{2}a{t}^2\\ & =\frac{1}{2}\times 2.459\times 10^{15}\times {\left(2.4\times 10^{-9}\right)}^2\\ & =7.1\times 10^{-3}\mathrm{m}\end{array}$ Step 4: Conclusion Since the vertical displacement of the electron as it leaves the plates ($7.1\times 10^{-3}$) is less than the distance from the centre line to the top plate ($7.5\times 10^{-3}$), $\therefore$ the electron emerges from between the plates.