Charge in Uniform Electric Fields
Australia's last CRT television broadcast ended 31 December 2013 (analogue switch-off). In the final generation of CRT sets, electrons were accelerated through 10,000 V: KE = QV = 1.6×10⁻¹⁹ × 10,000 = 1.6×10⁻¹⁵ J = 10 keV. Velocity on hitting the screen: v = √(2KE/m) = 5.9×10⁷ m/s = 0.2c.
An old television set fires electrons from a "gun" at the back of the tube, directing them to a screen at the front. The electrons are deflected horizontally and vertically by electric fields between charged plates to paint a picture on the screen.
Predict 1: How could charged plates deflect a beam of electrons without physically touching them?
Predict 2: If the electrons enter the region between the plates moving horizontally, will they follow a straight-line path, a circular path, or a parabolic path? Explain.
Warm-up — for a particle entering a uniform electric field perpendicular to the field direction, the trajectory is:
Know
- A uniform electric field produces a constant force on a charge: $F = qE$
- Acceleration of charge in uniform field: $a = qE/m$
- Energy method: $W = qV = \tfrac{1}{2}mv^2$
- Particles entering perpendicular to field follow parabolic paths
Understand
- The analogy between projectile motion under gravity and charged particle motion in a uniform electric field
- Why the path is parabolic (constant $v$ parallel, constant $a$ perpendicular)
Can Do
- Calculate the speed of a charge accelerated through a potential difference
- Calculate deflection of a charge entering perpendicular to a uniform field
- Apply kinematics equations to charged particle motion
Core Content
Inside a CRT television, an electron starts from rest at a hot cathode. Two metal plates — one positive, one negative — sit a few centimetres apart with 10,000 V between them. The electron accelerates from the negative plate toward the positive one, arriving at the screen with KE = 1.6×10⁻¹⁵ J and a speed of 5.9×10⁷ m/s. Every joule of electrical potential energy the field possessed has been converted to kinetic energy in the electron.
$W = qV = \tfrac{1}{2}mv^2$
$\Rightarrow v = \sqrt{\dfrac{2qV}{m}}$
Note: $q$ here is the magnitude of the charge. If the particle starts from rest, all the electrical potential energy converts to kinetic energy. For a particle not starting from rest, use the work-energy theorem: $W = \Delta KE$.
An electron ($q = 1.60 \times 10^{-19}\ \text{C}$, $m = 9.11 \times 10^{-31}\ \text{kg}$) is accelerated from rest through $V = 10\ \text{kV}$. Find its final speed.
- $W = qV = (1.60 \times 10^{-19})(10{,}000) = 1.6 \times 10^{-15}\ \text{J}$
- $v = \sqrt{2W/m} = \sqrt{2(1.6 \times 10^{-15})/(9.11 \times 10^{-31})} = \sqrt{3.51 \times 10^{15}} = 5.9 \times 10^7\ \text{m s}^{-1}$ ($\approx 0.2c$)
Energy conservation for a charge accelerated from rest: $qV = \tfrac{1}{2}mv^2$, so $v = \sqrt{2qV/m}$. For a particle not starting from rest, use $W = \Delta KE = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$.
Pause — copy the highlighted formula into your book before moving on.
A proton and an electron are both accelerated from rest through the same potential difference. Which statement is correct?
We just saw that a charge starting from rest converts all its electrical PE to KE. That raises a question: what happens when a pre-accelerated charge enters a uniform field sideways? This card answers it → the path is parabolic, identical in form to projectile motion.
When a charged particle enters a uniform electric field perpendicular to the field direction, it experiences a constant force perpendicular to its initial velocity. This is identical to projectile motion under gravity.
Two independent motions (analogous to projectile motion):
- Parallel to plates (horizontal): No force → constant velocity $v_0$. Displacement: $x = v_0 t$.
- Perpendicular to plates (vertical): Constant force $F = qE$ → constant acceleration $a = qE/m$. Displacement: $y = \tfrac{1}{2}at^2$.
$y = \tfrac{1}{2}at^2 = \tfrac{qE}{2m}\left(\dfrac{x}{v_0}\right)^2$
Parabolic ($y \propto x^2$) — identical mathematical form to projectile motion.
A charge entering a uniform field perpendicular to it follows a parabolic path. Horizontal: $x = v_0 t$ (constant speed). Vertical: $y = \tfrac{1}{2}at^2$ where $a = qE/m$. Combined: $y \propto x^2$. This is mathematically identical to projectile motion with $g \to a = qE/m$.
Add the highlighted principle to your notes before the check below.
A charged particle entering a uniform field perpendicular to the field follows a parabolic path.
A proton and an electron accelerated through the same voltage reach the same final speed.
An electron in a CRT is accelerated from rest through a potential difference of 5.0 kV.
- Calculate the kinetic energy gained by the electron.
- Calculate the final speed of the electron. ($m_e = 9.11 \times 10^{-31}\ \text{kg}$, $q = 1.60 \times 10^{-19}\ \text{C}$)
- If the voltage is doubled to 10 kV, by what factor does the speed increase?
An electron accelerated through 5.0 kV, then enters between deflection plates 4.0 cm long with a field of $2.0 \times 10^4\ \text{V m}^{-1}$. Which quantity is constant while the electron is between the plates?
An electron enters midway between two deflection plates, travelling horizontally at $v_0 = 3.0 \times 10^7\ \text{m s}^{-1}$. The plates are 5.0 cm long and separated by 2.0 cm, with a potential difference of 400 V across them.
- Calculate the electric field between the plates.
- Calculate the acceleration of the electron due to the field.
- Calculate the time the electron spends between the plates.
- Calculate the vertical deflection of the electron as it exits the plates.
Three statements correctly describe charged particle motion in a uniform electric field. Pick the odd one out.
An electron accelerated from rest through 2000 V: the kinetic energy gained is $3.2 \times 10^{-16}$ J. The speed reached (to 2 sig. figs.) is _____ × 10⁷ m s⁻¹.
Accelerated from rest
$qV = \tfrac{1}{2}mv^2$, so $v = \sqrt{2qV/m}$. Same KE for same $|q|$ and $V$; lighter particles are faster.
Parabolic deflection
Horizontal: $x = v_0 t$ (constant). Vertical: $y = \tfrac{1}{2}at^2$ where $a = qE/m$. Same as projectile motion.
Time between plates
$t = L/v_0$ where $L$ is the length of the plates. Use this to find vertical deflection.
Increasing deflection
Increase $E$ (increase $V$ or decrease $d$), decrease $v_0$, or increase plate length $L$. All increase $y$.
UnderstandBand 3(3 marks) 3. Explain the analogy between the motion of a charged particle entering a uniform electric field perpendicular to the field, and the projectile motion of a ball thrown horizontally under gravity. Identify what plays the role of gravity in the electric case.
ApplyBand 4(4 marks) 4. A proton ($m = 1.67 \times 10^{-27}\ \text{kg}$, $q = 1.60 \times 10^{-19}\ \text{C}$) is accelerated from rest through 20 kV and then enters a uniform field of $5.0 \times 10^4\ \text{V m}^{-1}$ perpendicular to its motion. Calculate: (a) the speed of the proton after acceleration; (b) the acceleration of the proton in the deflecting field; (c) the vertical deflection after travelling 3.0 cm horizontally through the field.
EvaluateBand 6(3 marks) 5. A CRT monitor designer wants to increase the vertical deflection of the electron beam. Evaluate two modifications that could be made to the deflection plates, explaining the physics behind each modification.
Show all answers
Activity 1 — Model Answers
- $KE = qV = (1.60 \times 10^{-19})(5000) = 8.0 \times 10^{-16}\ \text{J}$
- $v = \sqrt{2KE/m} = \sqrt{2(8.0 \times 10^{-16})/(9.11 \times 10^{-31})} = 4.2 \times 10^7\ \text{m s}^{-1}$
- $v \propto \sqrt{V}$. Doubling $V$ increases speed by $\sqrt{2} \approx 1.41$.
Activity 2 — Model Answers
- $E = V/d = 400/(2.0 \times 10^{-2}) = 2.0 \times 10^4\ \text{V m}^{-1}$
- $a = qE/m = (1.60 \times 10^{-19})(2.0 \times 10^4)/(9.11 \times 10^{-31}) = 3.51 \times 10^{15}\ \text{m s}^{-2}$
- $t = L/v_0 = (5.0 \times 10^{-2})/(3.0 \times 10^7) = 1.67 \times 10^{-9}\ \text{s}$
- $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(3.51 \times 10^{15})(1.67 \times 10^{-9})^2 = 4.9 \times 10^{-3}\ \text{m} = 4.9\ \text{mm}$
Short Answer — Model Answers
Q3 (3 marks): The horizontal direction has no force → constant horizontal velocity (same as horizontal throw under gravity where horizontal $v$ is constant). The vertical direction has a constant force ($F = qE$) perpendicular to the initial velocity → constant acceleration $a = qE/m$ (gravity plays this role in projectile motion). The combined result is a parabolic path, just as in projectile motion.
Q4 (4 marks): (a) $v = \sqrt{2qV/m} = \sqrt{2(1.60 \times 10^{-19})(20{,}000)/(1.67 \times 10^{-27})} = 1.96 \times 10^6\ \text{m s}^{-1}$. (b) $a = qE/m = (1.60 \times 10^{-19})(5.0 \times 10^4)/(1.67 \times 10^{-27}) = 4.79 \times 10^{12}\ \text{m s}^{-2}$. (c) $t = (3.0 \times 10^{-2})/(1.96 \times 10^6) = 1.53 \times 10^{-8}\ \text{s}$; $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(4.79 \times 10^{12})(1.53 \times 10^{-8})^2 = 5.6 \times 10^{-4}\ \text{m}$.
Q5 (3 marks): Mod 1: Increase the plate voltage $V$. This increases $E = V/d$, increasing $a = qE/m$, increasing deflection $y = \tfrac{1}{2}at^2$. Mod 2: Decrease the plate separation $d$. With the same $V$, $E = V/d$ increases → greater acceleration and deflection. (Could also mention lengthening the plates — increases $t$ in the field, increasing $y$.)
Five timed questions on charged particle motion in uniform fields.
⚔ Enter the arenaThe electrons in Australia's last CRT broadcast (31 December 2013) were accelerated through 10,000 V — giving KE = 1.6×10⁻¹⁵ J = 10 keV and v = 5.9×10⁷ m/s = 0.2c — then deflected by charged plates to paint the picture. Look back at your Think First predictions: charged plates deflect electrons through the electric force F = qE — no contact required. The path is parabolic because horizontal velocity is constant while vertical acceleration is constant — exactly the projectile motion analogy used in the CRT design.