Electric Potential and Charges in a Uniform Field – Electric Force and Electric Field (Senior 4 Physics) | Ganzaa

Electric Force and Electric Field

Lessons

Senior 4 · Physics

Electric Force and Electric Field

5 lessons

Lesson 3 of 5

Electric Potential and Charges in a Uniform Field

25 min

Drawing with electrons

Before flat screens, every television and oscilloscope drew its picture with a beam of electrons. Inside a glass tube, electrons boiled off a hot wire, were speeded up by a few thousand volts, and then were steered up, down and sideways by charged metal plates before striking a glowing screen. Two ideas make this possible: the potential difference that gives the electrons energy, and the uniform field between parallel plates that bends their path into a parabola, just like a thrown ball. In this lesson you define electric potential, link it to field strength, and calculate the speed and the deflection of charged particles.

Work, potential energy and potential

Moving a positive charge $q$ against an electric field takes work, just as lifting a mass against gravity does; the work is stored as electric potential energy. The electric potential $V$ at a point is the potential energy per unit charge there: the work done per unit positive charge in bringing a small test charge from infinitely far away (where $V = 0$ ) to that point,

V = \frac{W}{q} \qquad (1\,\mathrm{V} = 1\,\mathrm{J\,C^{-1}})

For a point charge $Q$ , adding up the work against the varying force $\frac{kQq}{r^{2}}$ from infinity to distance $r$ gives

V = \frac{kQ}{r} = \frac{Q}{4\pi\varepsilon_0 r}

Potential is a scalar: it has a sign but no direction, so the potential due to several charges is simply the algebraic sum $V = \sum \frac{kQ_i}{r_i}$ . Near a positive charge the potential is positive and falls with distance; near a negative charge it is negative. Two charges $q_{1}$ and $q_{2}$ a distance $r$ apart have potential energy $U = \frac{kq_1 q_2}{r}$ : positive (repulsion) for like charges, negative (bound together) for unlike charges.

The potential difference between two points is the work done per unit charge in moving a charge from one to the other, $W = q V$ . Surfaces on which the potential is the same everywhere are equipotentials; no work is done moving along one, so field lines always cross them at right angles (Figure 3.1).

Definition

The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point: $V = \frac{W}{q}$ , in volts ( $\mathrm{J\,C^{-1}}$ ). For a point charge $V = \frac{kQ}{r}$ . The electronvolt (eV) is the energy gained by an electron moving through a potential difference of 1 V: $1\,\mathrm{eV} = 1.602 \times 10^{-19}\,\mathrm{J}$ . Example: 0.10 m from a charge of $+1.0\,\mathrm{nC}$ , $V = \frac{8.99 \times 10^{9} \times 1.0 \times 10^{-9}}{0.10} = 90\,\mathrm{V}$ .

The uniform field between parallel plates

Between two parallel plates a distance $d$ apart, connected to a supply of voltage $V$ , the field is uniform (Lesson 2). Moving a charge $q$ from the negative plate to the positive plate against the constant force $q E$ takes work $W = q E d$ . But by definition $W = q V$ . So

qEd = qV \qquad\Rightarrow\qquad E = \frac{V}{d}

This shows that $1\,\mathrm{N\,C^{-1}} = 1\,\mathrm{V\,m^{-1}}$ . More generally, the field is the potential gradient, $E = -\frac{\mathrm{d}V}{\mathrm{d}x}$ : it points the way the potential falls fastest, and it is strong where equipotentials are close together.

Figure 3.1: The uniform field between parallel plates, with equally spaced equipotentials. The potential falls by the same amount for each equal step from the positive plate to the negative plate.

Worked example 1: plates at 500 V

Two parallel plates 2.0 cm apart are connected to a 500 V supply. Find the field between them, the force on an electron there, and the work done on the electron by the field as it crosses from the negative to the positive plate.

E = \frac{V}{d} = \frac{500}{0.020} = 2.5 \times 10^{4}\,\mathrm{V\,m^{-1}}, \qquad F = eE = 1.602 \times 10^{-19} \times 2.5 \times 10^{4} = 4.0 \times 10^{-15}\,\mathrm{N}

W = eV = 1.602 \times 10^{-19} \times 500 = 8.0 \times 10^{-17}\,\mathrm{J} = 500\,\mathrm{eV}

Worked example 2: potential from two charges

A point P is 0.10 m from a charge of $+3.0\,\mathrm{nC}$ and 0.20 m from a charge of $-1.0\,\mathrm{nC}$ . Find the potential at P, and the work needed to bring a charge of $+2.0\,\mathrm{nC}$ from infinity to P.

V = 8.99 \times 10^{9}\left(\frac{3.0 \times 10^{-9}}{0.10} + \frac{-1.0 \times 10^{-9}}{0.20}\right) = 8.99 \times 10^{9} \times 2.5 \times 10^{-8} = 225\,\mathrm{V}

W = qV = 2.0 \times 10^{-9} \times 225 = 4.5 \times 10^{-7}\,\mathrm{J}

No angles were needed: potentials add as plain numbers with their signs.

Charged particles in a uniform field

Along the field. A particle of charge $q$ and mass $m$ released in a uniform field feels a constant force $q E$ , so it has a constant acceleration $a = \frac{qE}{m}$ and moves like a body falling freely. If it starts from rest and moves through a potential difference $V$ , its gain in kinetic energy equals the work done on it:

\tfrac{1}{2} m v^{2} = qV \qquad\Rightarrow\qquad v = \sqrt{\frac{2qV}{m}}

Across the field. A particle entering the field at right angles with speed $v$ (Figure 3.2) behaves like a projectile launched horizontally (Unit 7): its velocity along the plates stays $v$ , while it accelerates across them with $a = \frac{qE}{m}$ . After travelling a distance $x$ along the plates, in time $t = \frac{x}{v}$ , its sideways displacement is

y = \tfrac{1}{2} a t^{2} = \frac{qE}{2mv^{2}}\,x^{2}

The path is a parabola. On leaving plates of length $L$ , its sideways velocity is $v_y = \frac{qEL}{mv}$ , and it then travels in a straight line at angle $\theta$ to its original direction, where $\tan\theta = \frac{v_y}{v} = \frac{qEL}{mv^{2}}$ . The weight of an electron or ion is negligible: for an electron, $\frac{eE}{m}$ in even a weak field is billions of times $g$ .

Figure 3.2: An electron crossing a uniform field: a parabola between the plates, a straight line after them.

Applications include the cathode-ray oscilloscope, where two pairs of plates steer the beam up–down and left–right; the inkjet printer, where tiny charged drops of ink are deflected by plates onto the paper; and the electrostatic precipitator, which charges dust particles in a factory chimney and pulls them onto collecting plates.

Worked example 3: an electron gun

Electrons start from rest and are accelerated through 2000 V. Find their kinetic energy and speed ( $m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$ ).

E_k = eV = 2000\,\mathrm{eV} = 2000 \times 1.602 \times 10^{-19} = 3.20 \times 10^{-16}\,\mathrm{J}

v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 3.20 \times 10^{-16}}{9.11 \times 10^{-31}}} = 2.65 \times 10^{7}\,\mathrm{m\,s^{-1}}

That is almost a tenth of the speed of light, gained in a few centimetres.

Worked example 4: deflecting the beam

The electrons of Worked example 3 enter, at right angles to the field, the gap between plates 4.0 cm long and 1.0 cm apart, with 100 V between them. Find the deflection when they leave the plates and the angle of the beam.

E = \frac{100}{0.010} = 1.0 \times 10^{4}\,\mathrm{V\,m^{-1}}, \qquad a = \frac{eE}{m} = \frac{1.602 \times 10^{-19} \times 1.0 \times 10^{4}}{9.11 \times 10^{-31}} = 1.76 \times 10^{15}\,\mathrm{m\,s^{-2}}

t = \frac{L}{v} = \frac{0.040}{2.65 \times 10^{7}} = 1.51 \times 10^{-9}\,\mathrm{s}, \qquad y = \tfrac{1}{2}at^{2} = \tfrac{1}{2} \times 1.76 \times 10^{15} \times (1.51 \times 10^{-9})^{2} = 2.0 \times 10^{-3}\,\mathrm{m}

v_y = at = 2.65 \times 10^{6}\,\mathrm{m\,s^{-1}}, \qquad \tan\theta = \frac{v_y}{v} = 0.100 \;\Rightarrow\; \theta = 5.7^\circ

The beam leaves the plates 2.0 mm off its original line; a screen 20 cm further on sees a spot about 22 mm off centre.

Warning

Common mistakes. (1) Treating potential as a vector, or field as a scalar: add potentials as signed numbers, fields as arrows. (2) Using $V = \frac{kQ}{r^{2}}$ : potential falls as $\frac{1}{r}$ , field as $\frac{1}{r^{2}}$ . (3) Using $E = \frac{V}{d}$ for the field of a point charge: it holds only in a uniform field. (4) Forgetting to convert eV to joules before using $\tfrac{1}{2}mv^{2}$ . (5) Using the gap between the plates instead of their length for the time spent between them.

Tip

Energy for speed, force for deflection. To find how fast a charge goes after crossing a potential difference, use energy: $qV = \tfrac{1}{2}mv^{2}$ . To find where it goes in a uniform field, use force and projectile motion: $a = \frac{qE}{m}$ , with $E = \frac{V}{d}$ .

Electron deflected by charged plates: projectile motion electrostatics problem (Zak's Lab). Watch how the force qE gives the electron a constant sideways acceleration, and how the projectile-motion equations give its deflection, as in Worked example 4.

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Practice

Two parallel plates 3.0 cm apart have a potential difference of 1500 V. Find the field between them.
Find the potential 5.0 cm and 10 cm from a charge of $+2.0\,\mathrm{nC}$ . How much work is done in moving a charge of $+1.0\,\mathrm{nC}$ from the 10 cm point to the 5.0 cm point?
Find the speed of an electron accelerated from rest through 100 V.
Find the electric potential energy of charges of $+1.0\,\mu\mathrm{C}$ and $-2.0\,\mu\mathrm{C}$ that are 0.30 m apart, and explain what its sign means.
A proton ( $m = 1.67 \times 10^{-27}\,\mathrm{kg}$ ) is accelerated from rest through 5.0 kV. Find its kinetic energy in eV and in joules, and its speed.
An electron moving at $3.0 \times 10^{7}\,\mathrm{m\,s^{-1}}$ enters, at right angles to the field, plates 5.0 cm long and 2.0 cm apart with 200 V between them. Find its deflection as it leaves the plates and the angle of its path.
An electron is accelerated from rest through a potential difference $V_{a}$ and then crosses plates of length $L$ with field $E$ . Show that its deflection on leaving is $y = \frac{EL^{2}}{4V_a}$ , and explain why this means every particle of the same charge sign follows the same path.
In an inkjet printer, drops of ink are given different charges before passing between deflecting plates. Analyse how the printer uses this to place each drop, and why uncharged drops are caught before the paper.

Show answersHide answers

$E = \frac{1500}{0.030} = 5.0 \times 10^{4}\,\mathrm{V\,m^{-1}}$ .
$V_{5} = \frac{8.99 \times 10^{9} \times 2.0 \times 10^{-9}}{0.050} = 360\,\mathrm{V}$ ; $V_{10} = 180\,\mathrm{V}$ ; $W = q\Delta V = 1.0 \times 10^{-9} \times 180 = 1.8 \times 10^{-7}\,\mathrm{J}$ .
$v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \times 100}{9.11 \times 10^{-31}}} = 5.9 \times 10^{6}\,\mathrm{m\,s^{-1}}$ .
$U = \frac{8.99 \times 10^{9} \times (1.0 \times 10^{-6})(-2.0 \times 10^{-6})}{0.30} = -0.060\,\mathrm{J}$ . Negative: the charges attract, and 0.060 J of work must be done to pull them infinitely far apart.
$E_k = 5.0\,\mathrm{keV} = 5000 \times 1.602 \times 10^{-19} = 8.0 \times 10^{-16}\,\mathrm{J}$ ; $v = \sqrt{\frac{2 \times 8.0 \times 10^{-16}}{1.67 \times 10^{-27}}} = 9.8 \times 10^{5}\,\mathrm{m\,s^{-1}}$ .
$E = 1.0 \times 10^{4}\,\mathrm{V\,m^{-1}}$ , $a = 1.76 \times 10^{15}\,\mathrm{m\,s^{-2}}$ , $t = 1.67 \times 10^{-9}\,\mathrm{s}$ ; $y = \tfrac{1}{2}at^{2} = 2.4\,\mathrm{mm}$ ; $v_y = 2.9 \times 10^{6}\,\mathrm{m\,s^{-1}}$ , $\theta = \tan^{-1}\frac{2.93}{30} = 5.6^\circ$ .
$eV_a = \tfrac{1}{2}mv^{2} \Rightarrow mv^{2} = 2eV_a$ ; $y = \frac{eEL^{2}}{2mv^{2}} = \frac{eEL^{2}}{4eV_a} = \frac{EL^{2}}{4V_a}$ . The charge and mass cancel, so electrons, and any ions accelerated through the same $V_{a}$ , are deflected by the same amount: the path depends only on the voltages and the plates.
All drops move at about the same speed, so the deflection $y = \frac{qEL^{2}}{2mv^{2}}$ is proportional to each drop's charge-to-mass ratio. The printer charges each drop by the amount that will send it to the right spot on the paper; drops that are not needed are left uncharged, travel straight on and are caught by a gutter and recycled.

Quick check

Key points

Potential $V = \frac{W}{q}$ (V = J/C), a scalar; point charge $V = \frac{kQ}{r}$ ; potentials add algebraically; $U = \frac{kq_1q_2}{r}$ .
Work in moving charge $q$ through potential difference $V$ : $W = q V$ ; $1\,\mathrm{eV} = 1.602 \times 10^{-19}\,\mathrm{J}$ .
Uniform field: $E = \frac{V}{d}$ ; in general $E = -\frac{\mathrm{d}V}{\mathrm{d}x}$ ; field lines cross equipotentials at right angles.
Accelerated from rest: $v = \sqrt{\frac{2qV}{m}}$ . Across a uniform field: a parabola, $y = \frac{qEx^{2}}{2mv^{2}}$ , $\tan\theta = \frac{qEL}{mv^{2}}$ .

Sources and credits

Photo: D9-10GH oscilloscope CRT tube by Mister rf, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Electron deflected by charged plates projectile motion electrostatics problem, Zak's Lab, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Drawing with electrons

2Work, potential energy and potential

3The uniform field between parallel plates

4Charged particles in a uniform field

Drawing with electrons

Work, potential energy and potential

The uniform field between parallel plates

Charged particles in a uniform field