Electricity and Magnetism 203-SN2-RE • CEGEP, Montreal

Revision sheet: electric charge, field and potential (203-SN2-RE)

Q: What is the difference between the electric field and the electric potential?

The field is a vector: it has a direction and adds by components. The potential is a scalar: it has no direction and adds like a signed number. The field falls off as the inverse square of the distance, the potential as the inverse of the distance. At one and the same point, one can be zero without the other being so.

Q: Where does the electric field of two charges vanish?

Between the two charges if they have the same sign, and outside, on the side of the weaker one, if they have opposite signs. The reasoning starts with the directions: between two opposite charges, the two fields point the same way and so can never cancel.

Q: Does zero flux mean the field is zero?

No. Zero flux only means that as many field lines enter the closed surface as leave it, so that it contains no net charge. A charge placed just outside creates a nonzero field at every point of the surface, while the total flux stays exactly zero.

Q: Why is the field zero inside a conductor?

Because the electrons there are free to move. If any field remained, they would move, and we would not be in equilibrium. They therefore redistribute themselves on the outer surface until they create a field that exactly cancels the outside field inside the metal. This is the principle of the Faraday cage.

Q: How do you calculate the energy of a system of three charges?

You add one term per pair of charges, that is three terms for three charges, and you keep the signs of the charges. Each pair is counted only once, since the energy belongs to the pair and not to each charge separately. A negative total energy signals a bound system.

Q: How do you find the speed of an electron accelerated by a voltage?

You equate the electric energy gained to the kinetic energy acquired: the charge times the potential difference equals half the mass times the speed squared. No distance is needed. For a few kilovolts, you get a few tens of millions of metres per second, about a tenth of the speed of light.

Electrostatics in 203-SN2-RE (formerly 203-NYB) introduces four quantities that look alike and are not handled the same way at all: force, field, potential and potential energy. Two are vectors, two are scalars, and every wrong turn costs a whole question.

This sheet collects what actually gets corrected in tutoring sessions: sloppy vector additions, the sign of the charge lost in the potential, Gauss's law applied without symmetry, zero flux mistaken for zero field, and the double counting of pairs in the energy of a system of charges. It is written for CEGEP students in the Science program (Sciences de la nature).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The field is a VECTOR, the potential a SCALAR. Almost every mark lost in this chapter comes from that single dividing line: you add components for one, signed numbers for the other.

This chapter is part of 203-SN2-RE, Electricity and Magnetism

The essentials

Charging: friction, contact, induction

• Charge is quantized, $q=ne$ with $e=1.6\times 10^{-19}$ C, and conserved: charging an object means moving electrons, never creating them.
• In a conductor, the free electrons move throughout the object; in an insulator, the charge stays where it was deposited.
• By induction, a charged object brought near a neutral conductor pushes its free electrons to the far face: the near face takes the opposite sign, and the conductor is attracted because that opposite charge is closer.

The negative rod pushes the free electrons to the right face of the sphere; the left face, which is closer, stays positive. Attraction wins: the neutral sphere is attracted.

A neutral object is always attracted by a charged object, never repelled: only repulsion proves that two objects are charged, and with the same sign.

Coulomb's law: a pair of forces

• $F=\dfrac{k|q_{1}||q_{2}|}{r^{2}}$ , with $r$ the distance between the centres: two charges of the same sign repel, two charges of opposite signs attract.
• Newton's third law: each charge feels a force of the same magnitude and opposite direction, whatever the difference between the two charges.
• The magnitude is computed with absolute values, the line of action is the line joining the two charges, and the direction is decided on the diagram.

Top, same sign: repulsion; bottom, opposite signs: attraction. The two arrows of a pair have the same length:

+4.0

μC and

-2.0

μC at

6.0

cm each feel

20

Doubling the distance divides the force by four; doubling only one of the two charges doubles the force, on BOTH charges at once.

Several charges: adding the forces tip to tail

• Each charge exerts its force as if it were alone: $F_{i}=\dfrac{k|q_{i}||q_{0}|}{r_{i}^{2}}$ , with ITS distance, along the line joining it to $q_{0}$ .
• You add VECTORS: $F_{x}=\sum F_{ix}$ and $F_{y}=\sum F_{iy}$ , then $F=\sqrt{F_{x}^{2}+F_{y}^{2}}$ and the direction from $\tan\theta=\dfrac{F_{y}}{F_{x}}$ .
• Adding magnitudes is correct only for forces along the same line and in the same direction.

q_{1}

repels

q_{0}

with

0.45

q_{2}

attracts it with

0.36

N. Placed tip to tail, to scale, these forces give the resultant in black:

0.69

N, not

0.81

N, tilted toward

q_{2}

Draw the tip-to-tail sum before calculating: the arrow you get gives the order of magnitude and the quadrant, and exposes a forgotten sign.

The dividing line: vector or scalar

• VECTORS, to be added by components: the force $\vec{F}$ and the field $\vec{E}$ . $E=\dfrac{k|q|}{r^{2}}$ with $k=9.0\times 10^{9}$ N·m $^{2}$ /C $^{2}$ , direction fixed by the SIGN of the source charge.
• SCALARS, to be added as signed numbers: the potential $V=\dfrac{kq}{r}$ and the energy $U=qV$ . The sign of $q$ goes straight into the value, with no components to project.
• $\vec{F}=q\vec{E}$ : if $q<0$ , the force is OPPOSITE to the field. This is the only way the sign of a test charge comes in.
• Watch the exponents: the field goes as $\dfrac{1}{r^{2}}$ , the potential as $\dfrac{1}{r}$ . Only one of these two formulas has a square.
• Work done by the field to move $q$ from A to B: $W=q(V_{A}-V_{B})$ ; work done by the external operator: the opposite.

At the midpoint

M

of two identical charges, the two fields are OPPOSITE and cancel, while the two potentials add: zero field and zero potential do not go together.

Before each calculation, write “vector” or “scalar” in the margin. That one word wins more marks than any extra formula you learn.

Field lines and equipotentials

• A field line starts on a positive charge and ends on a negative charge; the field is tangent to it at every point, and is strong where the lines crowd together.
• Two lines never cross: at a point, the field has only one direction.
• Equipotentials cut the field lines at right angles; along an equipotential, the electric force does no work.

Dipole: field lines in orange, from

+

-

; equipotentials in grey, perpendicular to the lines at every crossing. Both families crowd together where the field is strong.

On the perpendicular bisector of a dipole, $V=0$ everywhere, and yet the field there is nonzero and parallel to the axis: the value of $V$ says nothing about $\vec{E}$ .

The uniform field, and the energy that goes with it

• Between two parallel plates: $E=\dfrac{\Delta V}{d}$ , directed from the positive plate to the negative one, and uniform everywhere between the plates.
• The N/C and the V/m are the SAME unit: $1$ N/C $=1$ V/m.
• Charge accelerated through a potential difference, starting from rest: $\dfrac{1}{2}mv^{2}=|q|\,\Delta V$ . Mass and charge appear only there.
• In a uniform field, a charge follows a PARABOLA, exactly like a projectile: uniform motion in one direction, constant acceleration $a=\dfrac{|q|E}{m}$ in the other.
• Energy of a system of charges: one term $\dfrac{kq_{i}q_{j}}{r_{ij}}$ per PAIR, with the signs, each pair counted only once. $U<0$ means a bound system.

Between the plates, parallel and evenly spaced lines, from

+

-

: the field is uniform. The equipotentials, in grey, are parallel to the plates; the potential falls in the direction of the field.

Three charges give three pairs, four charges give six. Counting the pairs before calculating avoids the costliest error of the energy question.

Gauss and conductors: the results to know

• $\Phi_{E}=\dfrac{Q_{enc}}{\varepsilon_{0}}$ through a CLOSED surface, with $\varepsilon_{0}=8.85\times 10^{-12}$ C $^{2}$ /(N·m $^{2}$ ) and $k=\dfrac{1}{4\pi\varepsilon_{0}}$ .
• Charged sphere, seen from outside: it behaves like a point charge placed at its centre.
• Infinite line: $E=\dfrac{2k\lambda}{r}$ . Infinite plane: $E=\dfrac{\sigma}{2\varepsilon_{0}}$ , independent of distance. Parallel-plate capacitor: $E=\dfrac{\sigma}{\varepsilon_{0}}$ .
• Conductor in equilibrium: $E=0$ inside the metal, all the excess charge on the outer surface, field perpendicular to the surface.
• An empty cavity inside a conductor is shielded from the outside field: this is the Faraday cage.

Charge on the surface, zero field in the metal and perpendicular to the surface just outside. On the Gaussian surface

\Sigma

, the sphere acts like a point charge placed at its centre.

Gauss's law is always true, but it only lets you CALCULATE in three symmetries: spherical, cylindrical, planar. Anywhere else, you need superposition or integration.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Looking for the zero-field point BETWEEN two charges of opposite signs

the whole question, 6 marks

What not to write

“With $q_{1}=+9.0$ μC at $x=0$ and $q_{2}=-4.0$ μC at $x=3.0$ m, the field vanishes between the two, closer to the smaller charge.”

What to write

“Between the two, both fields point in the SAME direction and cannot cancel. The zero point is outside, on the side of the smaller charge: $\dfrac{9}{x^{2}}=\dfrac{4}{(x-3)^{2}}$ gives $x=9.0$ m.”

Between the two charges, both fields point to the right: their sum cannot vanish. The zero point is at

x=9.0

m, beyond the weaker charge. Arrows enlarged eighteen times at the zero point.

Why: The reasoning is done on the DIRECTIONS before the magnitudes. Two charges of the same sign cancel their fields between them; two charges of opposite signs cancel them only outside, beyond the weaker one.

2. Adding the magnitudes of the fields instead of their components

the whole question, 5 marks

What not to write

“The two charges create fields of $4.0\times 10^{4}$ N/C and $3.0\times 10^{4}$ N/C at P, so $E=7.0\times 10^{4}$ N/C.”

What to write

“The two fields make an angle with each other: I project. $E_{x}=\Sigma E_{ix}$ , $E_{y}=\Sigma E_{iy}$ , then $E=\sqrt{E_{x}^{2}+E_{y}^{2}}$ . Perpendicular, these two fields give $5.0\times 10^{4}$ N/C, not $7.0$ .”

The two perpendicular fields are placed tip to tail: the net field is the hypotenuse,

5.0\times 10^{4}

N/C, shorter than the sum

7.0\times 10^{4}

of the two lengths. Arrows to scale.

Why: Direct addition is valid only if the two fields are collinear and in the same direction. That is a special case, never the rule, and the problem always signals it with a figure where the charges are not aligned with the point.

3. Losing the sign of the charge in the potential

3 marks, and the energy calculation that follows

What not to write

“ $V=\dfrac{k|q_{1}|}{r_{1}}+\dfrac{k|q_{2}|}{r_{2}}$ , in absolute values as for the field.”

What to write

“ $V=\dfrac{kq_{1}}{r_{1}}+\dfrac{kq_{2}}{r_{2}}$ WITH the signs: a negative charge lowers the potential. The net potential can be zero, positive or negative.”

Why: Absolute values are for the field, because its direction already carries the information about the sign. The potential has no direction: its sign carries everything, and removing it amounts to erasing the data.

4. Believing that the force on a charge always follows the field

the whole question, the trajectory starts off on the wrong side

What not to write

“The field points to the right, so the electron placed there is pushed to the right.”

What to write

“ $\vec{F}=q\vec{E}$ and the electron has $q<0$ : the force is OPPOSITE to the field, so directed to the left.”

In the same field, the proton is pushed along the lines, the electron the opposite way:

\vec{F}=q\vec{E}

, and it is the sign of

q

that decides.

Why: The field is defined by the force on a POSITIVE test charge. For any negative charge, the relation reverses, and that is the case of the electron, that is, of half the problems in the course.

5. Counting the pairs twice in the energy of a system

the whole question, a factor 2 on the result

What not to write

“Three charges, so six terms: $U_{12}+U_{21}+U_{13}+U_{31}+U_{23}+U_{32}$ .”

What to write

“Three charges give THREE pairs: $U=\dfrac{kq_{1}q_{2}}{r_{12}}+\dfrac{kq_{1}q_{3}}{r_{13}}+\dfrac{kq_{2}q_{3}}{r_{23}}$ .”

Why: The energy belongs to the PAIR, not to each charge separately: the work to bring A closer to B is the same as to bring B closer to A, it is not two works. The number of pairs is $\dfrac{n(n-1)}{2}$ .

6. Concluding that the field is zero because the flux is

3 marks, and an absurd answer when there is an outside charge

What not to write

“The Gaussian surface contains no charge, so $\Phi=0$ , so $E=0$ over the whole surface.”

What to write

“ $\Phi=0$ means as many field lines enter as leave. The field itself is nonzero everywhere on the surface: the flux is an algebraic SUM, not a measure of the field.”

The charge is OUTSIDE the surface: every field line that enters eventually leaves, so the total flux is zero. The field, though, is nonzero at every point of this surface.

Why: Gauss's law gives the field only if the symmetry lets you take $E$ out of the integral. Without symmetry it stays true but mute, and that is precisely what the question wants you to say.

7. Writing the potential as $\dfrac{1}{r^{2}}$ or the field as $\dfrac{1}{r}$

2 marks per occurrence, and often three occurrences per paper

What not to write

“ $V=\dfrac{kq}{r^{2}}$ ”

What to write

“ $E=\dfrac{k|q|}{r^{2}}$ and $V=\dfrac{kq}{r}$ : the square belongs to the field, never to the potential.”

Why: The potential is the integral of the field along a displacement: integrating $\dfrac{1}{r^{2}}$ gives $\dfrac{1}{r}$ . The rule can be rebuilt in ten seconds rather than memorized, and it is checked by the units.

8. Forgetting that the field is zero INSIDE a conductor

the whole question, 4 marks

What not to write

“The metal sphere carries a charge $Q$ : at half the radius, the field is $\dfrac{kQ}{r^{2}}$ with $r$ equal to half the radius.”

What to write

“In a conductor in equilibrium, $E=0$ everywhere in the metal, whatever the charge carried: all the charge is on the outer surface. The formula holds only outside.”

Why: If the field were not zero, the free electrons would start moving and we would no longer be in equilibrium. The formula with $\dfrac{kQr}{R^{3}}$ holds for a uniformly charged INSULATING ball, never for a metal.

Which method to choose

Vector or scalar, so which addition

The quantity asked for, before any calculation

If you are asked for the FIELD created by several charges → vector: one magnitude per charge, then projection onto $x$ and onto $y$

Example: $E=\sqrt{E_{x}^{2}+E_{y}^{2}}$ and $\tan\theta=\dfrac{E_{y}}{E_{x}}$
If you are asked for the POTENTIAL at a point → scalar: algebraic sum of the $\dfrac{kq_{i}}{r_{i}}$ , signs included, with no projection at all

Example: two opposite charges at equal distances give $V=0$
If you are asked for the FORCE on a charge placed at a point → compute $\vec{E}$ , then $\vec{F}=q\vec{E}$ , and reverse the direction if $q<0$

Example: an electron in a field pointing right feels a force pointing left
If you are asked for the ENERGY of a charge placed at a point → $U=qV$ , a single product, with both signs

Example: $q<0$ in a $V>0$ gives $U<0$
If you are asked for the ASSEMBLY ENERGY of a system → a sum over PAIRS, $\dfrac{n(n-1)}{2}$ terms

Example: three charges give three terms
If you are asked for a WORK or a final speed → conservation of energy: $\dfrac{1}{2}mv^{2}=|q|\Delta V$ , never the force or the acceleration

Example: an electron accelerated through $2.00$ kV reaches $2.65\times 10^{7}$ m/s

A question containing the word “speed” or “work” is handled by energy, so by scalars. A question containing “direction” or “angle” is handled by the field, so by components. The wording of the problem decides.

Which method to find a field

The geometry of the charge distribution

If a single point charge → $E=\dfrac{k|q|}{r^{2}}$ , direction according to the sign

Example: no integral, no theorem
If two, three or four point charges → superposition by components

Example: net field at the vertex of a triangle
If SPHERICAL, CYLINDRICAL or PLANAR symmetry → Gauss's law, with the surface matched to the symmetry

Example: sphere, infinite line, infinite plane, parallel-plate capacitor
If a ring or a rod, on its axis → integrated formula: $E=\dfrac{kQx}{(x^{2}+a^{2})^{3/2}}$ for the ring

Example: a ring gives a ZERO field at its centre, by symmetry
If between two parallel plates → $E=\dfrac{\Delta V}{d}$ , uniform

Example: $60$ V across $1.5$ cm gives $4.0\times 10^{3}$ V/m
If inside the metal of a conductor in equilibrium → $E=0$ , no calculation

Example: Faraday cage, shielded cavity

Gauss's law remains true without symmetry, but it no longer lets you calculate: you cannot take $E$ out of the integral. A Gauss question with no announced symmetry therefore expects an answer about the FLUX, not about the field.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Applying Gauss's law, written up for the marker

When to use it: The problem announces a sphere, a line, a cylinder or a plane, and asks for the field at a given distance.

1 Name the symmetry and deduce the direction of the field: “the distribution has spherical symmetry, so $\vec{E}$ is radial and its magnitude depends only on $r$ .”
2 Choose the closed surface that matches this symmetry, and say so: sphere of radius $r$ , coaxial cylinder of length $L$ , box passing through the plane.
3 Justify taking $E$ out of the integral: “ $\vec{E}$ is constant in magnitude and parallel to $d\vec{A}$ over the whole side surface”, and point out the faces where the flux is zero.
4 Write $\Phi=EA=\dfrac{Q_{enc}}{\varepsilon_{0}}$ , spelling out $A$ and above all $Q_{enc}$ , the charge ACTUALLY contained in the chosen surface.
5 Solve for $E$ , then check the limiting case: from very far away, the distribution must behave like a point charge.

Concluding sentence

“By spherical symmetry, $\vec{E}$ is radial and of constant magnitude on the sphere of radius $r$ . The flux is therefore $E\cdot 4\pi r^{2}=\dfrac{Q_{enc}}{\varepsilon_{0}}$ , hence $E=\dfrac{Q_{enc}}{4\pi\varepsilon_{0}r^{2}}=\dfrac{kQ_{enc}}{r^{2}}$ .”

The trap: Putting the TOTAL charge in $Q_{enc}$ when the Gaussian surface contains only part of it. Inside a uniformly charged insulating ball, $Q_{enc}$ is only the fraction $\dfrac{r^{3}}{R^{3}}$ of the charge.

Marking: Often 1 mark for stating the symmetry, 1 for the choice of surface, 2 for the flux correctly written, 1 for $Q_{enc}$ , 1 for solving for $E$ .

Accelerating a charge through a potential difference

When to use it: The problem asks for a final speed, an energy or a flight time after a passage between two electrodes.

1 Choose energy rather than force: the question asks for neither direction nor duration, so there is no reason to compute an acceleration.
2 Write the conservation law: the potential energy lost becomes kinetic energy, $|q|\Delta V=\dfrac{1}{2}mv^{2}-\dfrac{1}{2}mv_{0}^{2}$ .
3 Check the DIRECTION: a charge is accelerated only if it moves toward the potential that suits it, an electron from low potential to high potential.
4 Solve for $v=\sqrt{\dfrac{2|q|\Delta V}{m}}$ and check the order of magnitude, a few $10^{7}$ m/s for an electron under a few kilovolts.

Concluding sentence

“The electron starts from rest and crosses a potential difference of $2.00$ kV: $\dfrac{1}{2}mv^{2}=e\Delta V$ , so $v=\sqrt{\dfrac{2(1.6\times 10^{-19})(2000)}{9.11\times 10^{-31}}}=2.65\times 10^{7}$ m/s.”

The trap: Going through the force and the acceleration, which forces you to know the distance between the electrodes. Energy needs only the potential difference, and that is why the problem does not give this distance.

Marking: 2 marks for the energy equation set up, 1 for the substitution, 1 for the result with its unit.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The word “vector” or “scalar” written in the margin

Go back over each line of your paper and check that fields are projected there and that potentials are added directly. A square root of a sum of squares applied to potentials is an immediate error.

Writing $V=\sqrt{V_{1}^{2}+V_{2}^{2}}$ : the potential has no components.
The exponents of $r$

Check by units: the field is in N/C, so in $\dfrac{1}{r^{2}}$ ; the potential is in volts, so in $\dfrac{1}{r}$ . The ratio $\dfrac{V}{E}$ must have the dimension of a length.

Getting a $\dfrac{V}{E}$ that comes out in square metres: one of the two exponents is wrong.
The sign of the result, read as information

A negative potential, a negative energy or a negative work are not errors: they respectively say a dominant negative charge, a bound system, and a displacement against the force.

Getting $U=-3.2\times 10^{-18}$ J for an electron near a proton is perfectly normal: the system is bound.
The order of magnitude of an electron's speed

Under a few kilovolts, an electron reaches a few $10^{7}$ m/s, about a tenth of the speed of light. Any value above $3\times 10^{8}$ m/s is impossible.

Getting $v=2.6\times 10^{9}$ m/s: the mass of the electron was taken for that of the proton, or the exponent is wrong.
The number of pairs in a system's energy

Count the pairs before calculating: $3$ charges give $3$ terms, $4$ give $6$ . Write this number in the margin, then count the terms actually present.

Six terms for three charges: each pair was counted twice, the energy is doubled.

The typical problem, taken apart

The electron gun: accelerate, then deflect

An electron starts from rest and is accelerated through a potential difference $\Delta V_{acc}=2.00$ kV. It then enters, parallel to the plates and midway between them, a parallel-plate capacitor of length $L=6.00$ cm whose plates are $d=1.50$ cm apart and have a potential difference $\Delta V_{p}=60.0$ V.

Find the entry speed, then the vertical deflection at the exit of the plates, and check that the electron does not hit a plate. Take $e=1.6\times 10^{-19}$ C and $m=9.11\times 10^{-31}$ kg.

Two stages, two kinds of physics: the field accelerates the electron ALONG its motion in the gun, then PERPENDICULAR to it between the plates, hence the parabola. The deflection is exaggerated here, it is only

1.80

mm.

Step 1

Entry speed, by energy: $e\Delta V_{acc}=\dfrac{1}{2}mv^{2}$ , so $v=\sqrt{\dfrac{2(1.6\times 10^{-19})(2000)}{9.11\times 10^{-31}}}=2.65\times 10^{7}$ m/s.

Why

Energy avoids the geometry of the gun altogether: neither the distance between cathode and anode nor the shape of the field comes in, which explains why the problem does not give them.

Step 2

Field between the plates: $E=\dfrac{\Delta V_{p}}{d}=\dfrac{60.0}{0.0150}=4.00\times 10^{3}$ V/m, directed from the positive plate to the negative one.

Why

This is the only formula in the chapter where the distance between the plates matters. The electron, being negatively charged, is attracted toward the POSITIVE plate, so in the direction opposite to the field.

Step 3

Transverse acceleration: $a=\dfrac{eE}{m}=\dfrac{(1.6\times 10^{-19})(4.00\times 10^{3})}{9.11\times 10^{-31}}=7.03\times 10^{14}$ m/s $^{2}$ .

Why

The weight of the electron, of order $10^{-29}$ N, is negligible compared with the electric force, of order $10^{-16}$ N: this is why gravity never appears in this type of problem.

Step 4

Transit time: the horizontal component of the velocity is UNCHANGED, so $t=\dfrac{L}{v}=\dfrac{0.0600}{2.65\times 10^{7}}=2.26\times 10^{-9}$ s.

Why

Exactly as for a projectile: the horizontal motion stays uniform because the field has no component in that direction. This decoupling is what makes the problem solvable.

Step 5

Vertical deflection: $y=\dfrac{1}{2}at^{2}=\dfrac{1}{2}(7.03\times 10^{14})(2.26\times 10^{-9})^{2}=1.80\times 10^{-3}$ m, that is $1.80$ mm.

Why

The electron starts midway, so it has $\dfrac{d}{2}=7.50$ mm before touching the plate. Since $1.80<7.50$ , it does leave the capacitor: this comparison is part of the answer.

Step 6

Check with the compact formula: $y=\dfrac{L^{2}\Delta V_{p}}{4d\,\Delta V_{acc}}=\dfrac{(0.0600)^{2}(60.0)}{4(0.0150)(2000)}=1.80\times 10^{-3}$ m.

Why

Neither $e$ nor $m$ appears in it: they cancelled. The check is therefore independent of the two constants used above, and that is what makes it a real check.

Step 7

Exit angle: $\tan\theta=\dfrac{v_{y}}{v_{x}}=\dfrac{at}{v}=\dfrac{1.59\times 10^{6}}{2.65\times 10^{7}}=0.0600$ , so $\theta=3.43^{\circ}$ .

Why

The angle stays small, which confirms the small deflection. A value of several tens of degrees would signal an exponent error on the field or the speed.

The conclusion, written out

“The electron enters the capacitor at $2.65\times 10^{7}$ m/s. The field between the plates is $4.00\times 10^{3}$ V/m and gives it a transverse acceleration of $7.03\times 10^{14}$ m/s $^{2}$ for $2.26$ ns, hence a deflection of $1.80$ mm. Since $1.80$ mm is less than the half-gap of $7.50$ mm, the electron leaves the plates, at an angle of $3.43^{\circ}$ .”

The classic mistake on this problem: Computing the deflection with the final vertical speed instead of the average, that is, writing $y=v_{y}t$ instead of $y=\dfrac{1}{2}at^{2}$ . The result is exactly doubled, and nothing in the units gives it away.

Learn by heart

• $E=\dfrac{k|q|}{r^{2}}$ is a vector; $V=\dfrac{kq}{r}$ is a SIGNED scalar. The square belongs to the field.
• $\vec{F}=q\vec{E}$ : for $q<0$ , the force is opposite to the field.
• Two opposite charges: zero field OUTSIDE, on the side of the weaker one. Two charges of the same sign: zero field BETWEEN them.
• Energy of a system: one term per PAIR, $\dfrac{n(n-1)}{2}$ terms, signs included.
• $\Phi=\dfrac{Q_{enc}}{\varepsilon_{0}}$ : zero flux does not mean zero field.
• Conductor in equilibrium: $E=0$ in the metal, charge on the outer surface.
• Uniform field: $E=\dfrac{\Delta V}{d}$ , and $1$ N/C $=1$ V/m.
• Charge accelerated from rest: $\dfrac{1}{2}mv^{2}=|q|\Delta V$ , a few $10^{7}$ m/s for an electron under a few kilovolts.

Frequently asked questions

What is the difference between the electric field and the electric potential?

The field is a vector: it has a direction and adds by components. The potential is a scalar: it has no direction and adds like a signed number. The field falls off as the inverse square of the distance, the potential as the inverse of the distance. At one and the same point, one can be zero without the other being so.

Where does the electric field of two charges vanish?

Between the two charges if they have the same sign, and outside, on the side of the weaker one, if they have opposite signs. The reasoning starts with the directions: between two opposite charges, the two fields point the same way and so can never cancel.

Does zero flux mean the field is zero?

No. Zero flux only means that as many field lines enter the closed surface as leave it, so that it contains no net charge. A charge placed just outside creates a nonzero field at every point of the surface, while the total flux stays exactly zero.

Why is the field zero inside a conductor?

Because the electrons there are free to move. If any field remained, they would move, and we would not be in equilibrium. They therefore redistribute themselves on the outer surface until they create a field that exactly cancels the outside field inside the metal. This is the principle of the Faraday cage.

How do you calculate the energy of a system of three charges?

You add one term per pair of charges, that is three terms for three charges, and you keep the signs of the charges. Each pair is counted only once, since the energy belongs to the pair and not to each charge separately. A negative total energy signals a bound system.

How do you find the speed of an electron accelerated by a voltage?

You equate the electric energy gained to the kinetic energy acquired: the charge times the potential difference equals half the mass times the speed squared. No distance is needed. For a few kilovolts, you get a few tens of millions of metres per second, about a tenth of the speed of light.

Practise it

Corrected exercises: Electric charge, field and potential, 203-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

13 corrected exercises
130 points
195 minutes

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys2-charge-field-potential. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: electric charge, field and potential (203-SN2-RE)

The essentials

Charging: friction, contact, induction

Coulomb's law: a pair of forces

Several charges: adding the forces tip to tail

The dividing line: vector or scalar

Field lines and equipotentials

The uniform field, and the energy that goes with it

Gauss and conductors: the results to know

The mistakes that cost marks

1. Looking for the zero-field point BETWEEN two charges of opposite signs

2. Adding the magnitudes of the fields instead of their components

3. Losing the sign of the charge in the potential

4. Believing that the force on a charge always follows the field

5. Counting the pairs twice in the energy of a system

6. Concluding that the field is zero because the flux is

7. Writing the potential as 1r2\dfrac{1}{r^{2}}r21​ or the field as 1r\dfrac{1}{r}r1​

8. Forgetting that the field is zero INSIDE a conductor

Which method to choose

Vector or scalar, so which addition

Which method to find a field

How the answer is expected to be written

Applying Gauss's law, written up for the marker

Accelerating a charge through a potential difference

Check before you hand in

The typical problem, taken apart

The electron gun: accelerate, then deflect

Learn by heart

Frequently asked questions

What is the difference between the electric field and the electric potential?

Where does the electric field of two charges vanish?

Does zero flux mean the field is zero?

Why is the field zero inside a conductor?

How do you calculate the energy of a system of three charges?

How do you find the speed of an electron accelerated by a voltage?

Corrected exercises: Electric charge, field and potential, 203-SN2-RE

See also

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7. Writing the potential as $\dfrac{1}{r^{2}}$ or the field as $\dfrac{1}{r}$