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

Revision sheet: Gauss's law (203-SN2-RE)

This sheet does not redo the course on Gauss's law, it lists what costs marks on test day. The law has a puzzling property: it is universally true and almost always unusable. Its validity assumes no symmetry, but it only gives the field if a symmetry provides a surface on which the magnitude of $\vec{E}$ is constant and its direction known.

It is written for CEGEP students in the Natural Science program (Sciences de la nature) taking 203-SN2-RE (formerly 203-NYB), and it also serves as a review in first-year university physics, where the law has exactly the same place: the first tool that replaces an integral by an argument from symmetry.

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

Gauss's law is always TRUE, it is not always USEFUL: it is the symmetry, and nothing else, that decides whether there is a surface on which $E$ can be pulled out of the integral.

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

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Electric flux
Gauss's law
Gaussian surface
Spherical symmetry
Cylindrical symmetry
Planar symmetry
Volume charge density
Conductor in electrostatic equilibrium
Electrostatic shielding
Principle of superposition
Electrostatic pressure
Solid angle

The essentials

Flux: what passes through a surface

• Uniform field through a flat surface of area $A$ : $\Phi_{E}=EA\cos\theta$ , where $\theta$ is the angle between $\vec{E}$ and the normal $\vec{n}$ to the surface.
• Closed surface: $\vec{n}$ points OUTWARD; the flux is positive where the field leaves and negative where it enters.
• Flux counts the field lines that pass through: zero if $\vec{E}$ is parallel to the surface, greatest if it is perpendicular to it.

Seen edge-on, the normal

\vec{n}

of the surface makes the angle

\theta

with

\vec{E}

: only the component of

\vec{E}

along

\vec{n}

passes through,

\Phi_{E}=EA\cos\theta

, zero if the surface is parallel to the field.

The angle is measured from the NORMAL, not from the surface: confusing the two swaps cosine and sine.

The law, and the only three symmetries that make it useful

• $\displaystyle\oint\vec{E}\cdot d\vec{A}=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$ , with the ENCLOSED charge only. Outside charges create a field on the surface but a zero net flux: what goes in comes out.
• Spherical: $E(4\pi r^{2})=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$ . Infinite cylindrical: $E(2\pi rL)=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$ , the two end caps not contributing. Infinite planar: pillbox through the plane.
• SOLID uniformly charged sphere: $E=\dfrac{kQr}{R^{3}}$ inside, $E=\dfrac{kQ}{r^{2}}$ outside. The field grows linearly, reaches a maximum at $r=R$ , then decreases.
• Conductor in equilibrium: $E=0$ in the metal, $E=\dfrac{\sigma}{\varepsilon_{0}}$ just outside, that is DOUBLE an insulating sheet of the same $\sigma$ , because the flux leaves on one side only.

In the insulating sphere the field GROWS with

r

, because the enclosed charge grows as

r^{3}

while the surface only grows as

r^{2}

; it peaks at the surface, then decreases.

What the law cannot do

• It NEVER distinguishes two distributions of the same total charge seen from outside: hollow sphere, solid sphere, conductor or point charge give exactly the same field for $r>R$ .
• It gives nothing without symmetry: off-centre cavity, charged cube, finite rod. You then go through SUPERPOSITION, writing the object as a sum of symmetric pieces, real or fictitious.
• NON-uniform density: no memorized formula applies, you must integrate $Q_{\text{enc}}=\displaystyle\int_{0}^{r}\rho(r')4\pi r'^{2}dr'$ before applying the law.
• Off-centre spherical cavity in a uniform sphere: the interior field is UNIFORM and equals $\vec{E}=\dfrac{\rho\vec{d}}{3\varepsilon_{0}}$ , where $\vec{d}$ joins the two centres. Neither the radius of the cavity nor that of the sphere appears in it.

In the off-centre cavity, the field is the same everywhere, parallel to

\overrightarrow{OO'}

\dfrac{\rho d}{3\varepsilon_{0}}=1.58\times 10^{4}

V/m for

d=7.00

cm. No radius enters.

Hollow conductor: the charges that appear

• A Gaussian surface drawn INSIDE the metal sees $E=0$ , hence a zero flux, hence a zero enclosed charge: the wall of a cavity carries the opposite of the charge it surrounds.
• The outer surface carries the rest, by conservation: the total charge of the conductor minus that of the inner wall.
• Seen from outside, only the charge of the outer surface counts: the position of the charge in the cavity cannot be seen.

\Sigma

, drawn in the metal where

E=0

, encloses zero charge: the inner wall carries

-q

(blue signs), the outer surface

+q

(red signs). Outside, everything happens as for

q

alone.

Always in this order: Gauss in the metal, then conservation of the total charge. This is the reasoning behind Faraday's ice pail and the charged shell.

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. A cube is placed in a uniform electric field of $350$ N/C. The total flux through its surface is zero.

3 marks, and it is the number one error of the chapter

What not to write

The flux is zero, so the electric field is zero at every point of the cube's surface.

What to write

A zero flux only means that the enclosed charge is zero: what enters through one face leaves through the opposite face. The field is $350$ N/C everywhere.

As many field lines enter through the left face as leave through the right face: the balance is zero although the field is

350

N/C at every point. The normals

\vec{n}

point outward.

Why: Flux is a BALANCE between what goes in and what comes out, not a measure of what passes through. The converse is true: if the field is zero everywhere on the surface, so is the flux. The minimal counterexample to be able to cite is precisely this cube, and it is enough to answer any question of this kind.

2. A charge of $50$ nC is placed $1.00$ cm outside one face of the cube.

2 marks on the reasoning question

What not to write

This charge being outside, it creates no field on the surface of the cube, which is why it does not appear in the law.

What to write

It creates an enormous field on that face, $4.49\times 10^{6}$ N/C at one centimetre. What is zero is its contribution to the NET FLUX: its lines cross the surface from side to side.

The charge is outside the cube, very close to one face: its lines enter through that face and leave through the others. The net flux is zero, yet the field on the face is intense.

Why: The conclusion of the false statement is correct, but its justification is not, and a marker penalizes that. The law does not say that outside charges have no effect, it says that their flux balance is zero. This is an essential difference as soon as you are asked for the field at a point rather than the flux.

3. An insulating sphere of radius $R$ carries a charge density $\rho(r)=\rho_{0}\dfrac{r}{R}$ , increasing outward. Calculate the field at $r=\dfrac{R}{2}$ .

The whole question, 4 marks

What not to write

$E=\dfrac{kQr}{R^{3}}$ with the total charge, which gives $4.24\times 10^{3}$ V/m.

What to write

I first calculate $Q_{\text{enc}}(r)=\displaystyle\int_{0}^{r}\rho_{0}\dfrac{r'}{R}4\pi r'^{2}dr'=\dfrac{\pi\rho_{0}r^{4}}{R}$ , then $E=\dfrac{k\pi\rho_{0}r^{2}}{R}=2.12\times 10^{3}$ V/m.

Why: The formula $\dfrac{kQr}{R^{3}}$ assumes that the enclosed charge grows as $r^{3}$ , which is true only for a UNIFORM density. Here it grows as $r^{4}$ , the matter being concentrated toward the outside, so there is less charge than expected at the centre: the result is off by exactly a factor of 2 at $r=\dfrac{R}{2}$ . Nothing outside gives the difference away, both distributions giving the same field there.

4. A conducting plate carries $\sigma=8.00$ nC/m $^{2}$ on each of its faces. Calculate the field just outside.

2 marks, with a factor 2 that contaminates everything after it

What not to write

$E=\dfrac{\sigma}{2\varepsilon_{0}}=452$ N/C, as for any charged plane.

What to write

$E=\dfrac{\sigma}{\varepsilon_{0}}=904$ N/C, double, because the Gaussian pillbox has one end cap INSIDE the metal where the field is zero: the flux leaves on one side only.

Sheet: the flux leaves through BOTH end caps,

\dfrac{\sigma}{2\varepsilon_{0}}=452

N/C. Conductor: one end cap is in the metal, where

E=0

, everything leaves through the other,

\dfrac{\sigma}{\varepsilon_{0}}=904

N/C.

Why: Both formulas are right, they simply do not refer to the same object. Insulating sheet: the pillbox extends on both sides, the flux leaves through TWO end caps, hence the factor 2 in the denominator. Conductor: only one end cap is in vacuum, the other is in the metal where the field is zero. Another equivalent reading: a conductor carries $\sigma$ on its TWO faces, and two superposed sheets do give $\dfrac{\sigma}{\varepsilon_{0}}$ .

5. Compare the outside field of a hollow sphere and a solid sphere carrying the same charge.

2 marks, and a missed chance to save time

What not to write

They differ, since the charge is not distributed in the same way inside.

What to write

They are STRICTLY identical: an outside spherical Gaussian surface encloses the same charge in both cases, with the same symmetry, so $E=\dfrac{kQ}{r^{2}}$ on both sides.

Same charge, same Gaussian surface: the hollow sphere (charge on the surface) and the solid sphere (charge in the volume) give the same field

\dfrac{kQ}{r^{2}}

on the dashed surface. The arrows are identical.

Why: Seen from outside, any distribution with spherical symmetry is indistinguishable from an equal point charge placed at the centre. This is the strength of the law, and also its limit: it cannot be used to probe an internal structure from outside. It is for the gravitational version of this result that Newton had to prove that a planet attracts as if all its mass were at its centre.

6. You are asked for the field in a spherical cavity hollowed out away from the centre of a uniformly charged ball.

The whole question, often 5 marks at the end of a test

What not to write

Gauss's law applied to a small sphere in the cavity gives a zero field, since it encloses no charge.

What to write

The field is not zero: it is UNIFORM and equals $\dfrac{\rho d}{3\varepsilon_{0}}$ . It is obtained by superposing a ball of density $+\rho$ and a ball of density $-\rho$ filling the cavity.

Why: The law remains true, it only says that the FLUX is zero, which does not imply that the field is, there being no symmetry on that surface. Superposition unlocks everything: by vector addition, the observation point DISAPPEARS from the result, which proves the uniformity. Neither the radius of the cavity nor that of the ball appears, only the distance between the centres.

Which method to choose

Which Gaussian surface, according to the symmetry

Before any calculation, look for the invariances of the distribution: they alone say whether there is a surface on which $E$ is constant.

The three useful Gaussian surfaces, in cyan: concentric sphere, coaxial cylinder, pillbox through the plane. On each,

\vec{E}

is perpendicular to the surface with constant magnitude, or parallel to it.

If SPHERICAL symmetry: point charge, ball, shell → Concentric sphere of radius $r$ . $E(4\pi r^{2})=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$ . Treat $r<R$ and $r>R$ separately, then CHECK that the two expressions agree at $r=R$ .

Example: Ball of $45.0$ nC and $10.0$ cm: $1.62\times 10^{4}$ V/m at $4.00$ cm, $4.04\times 10^{4}$ V/m at the surface.
If Infinite CYLINDRICAL symmetry: wire, cable, charged cylinder → Coaxial cylinder of radius $r$ and length $L$ . The two end caps do not contribute, the field being parallel to their surface, and $L$ cancels.

Example: Coaxial cable of density $3.20$ nC/m: $E=1.92\times 10^{4}$ V/m at $3.00$ mm, and $50.6$ pF/m.
If Infinite PLANAR symmetry: sheet or plate → Cylindrical pillbox through the plane. $E=\dfrac{\sigma}{2\varepsilon_{0}}$ for an insulating sheet, $E=\dfrac{\sigma}{\varepsilon_{0}}$ at the surface of a conductor.

Example: Two planes of opposite signs: $\dfrac{\sigma}{\varepsilon_{0}}$ between them, zero outside.
If NON-uniform density but symmetry preserved → The law still applies, but you must INTEGRATE the density to get $Q_{\text{enc}}(r)$ before using it. No memorized formula fits.

Example: $\rho\propto r$ : $Q_{\text{enc}}\propto r^{4}$ and $E\propto r^{2}$ , instead of $r^{3}$ and $r$ .
If NO symmetry: off-centre cavity, finite object → Superposition: write the object as a sum of symmetric pieces, possibly fictitious and of negative density. The law then serves on each piece, not on the whole.

Example: Off-centre cavity of $7.00$ cm in $\rho=6.00$ μC/m $^{3}$ : uniform field of $1.58\times 10^{4}$ V/m.

A CONDUCTOR simplifies everything: the field is zero in it, so a Gaussian surface drawn in the metal always encloses zero charge. This two-step reasoning, Gauss in the metal then conservation of the total charge, gives the charges on the two surfaces of a shell.

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 properly

When to use it: As soon as a distribution has one of the three symmetries and you are asked for the field.

1 State the symmetry and deduce the DIRECTION of the field and the variables its magnitude depends on.
2 Choose the Gaussian surface and say why: the magnitude of $\vec{E}$ is constant on it, and its direction is known at every point.
3 Calculate the flux, distinguishing the parts of the surface where $\vec{E}$ is perpendicular to it from those where it is parallel and which do not contribute.
4 Calculate the ENCLOSED charge, distinguishing the regions $r<R$ and $r>R$ .
5 Solve, then CHECK the continuity at the matching point and the behaviour at infinity.

Concluding sentence

“The distribution is invariant under rotation about the axis and under translation along it, so $\vec{E}$ is radial and its magnitude depends only on $r$ : I choose a coaxial cylinder of radius $r$ and length $L$ .”

The trap: Writing $E=\dfrac{Q}{4\pi\varepsilon_{0}r^{2}}$ directly without having justified the symmetry: the marker cannot tell an application of the law from a copied formula, and half of the marks are precisely for the justification.

Marking: 2 marks for the symmetry and the justified choice of surface, 1 mark for the flux, 1 mark for the enclosed charge, 1 mark for the verified matching.

Check before you hand in

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

The matching at $r=R$

Evaluate the two expressions, inside and outside, exactly at $r=R$ . They must give the same number: it is the mandatory check for any field defined piecewise.

$\dfrac{kQR}{R^{3}}$ and $\dfrac{kQ}{R^{2}}$ both give $4.04\times 10^{4}$ V/m: the match is good.
Behaviour far from the object

Let $r$ tend to infinity: the field must decrease as $\dfrac{1}{r^{2}}$ for a bounded distribution, as $\dfrac{1}{r}$ for an infinite cylinder, and stay CONSTANT for an infinite plane.

A field that would decrease as $\dfrac{1}{r^{3}}$ far from a charged ball signals an error in the enclosed charge.
The field is zero in the metal

On any diagram containing a conductor, mark $E=0$ in its thickness before starting, then deduce the surface charges by Gauss.

Neutral shell around $+12.0$ nC: $-12.0$ nC on the inner face, $+12.0$ nC on the outer face.
Units and order of magnitude

A laboratory electrostatic field ranges from a few volts per metre to a few megavolts per metre, air breaking down at $3.00\times 10^{6}$ V/m. Beyond that, the situation described is physically impossible.

A field of $10^{9}$ V/m reported for a sphere carrying a few nanocoulombs signals a power-of-ten error.

The typical problem, taken apart

The solid insulating sphere, its two regions and their matching

An INSULATING sphere of radius $R=10.0$ cm carries $Q=45.0$ nC spread uniformly through its VOLUME. Find the field at $r=4.00$ cm, at the surface and at $r=25.0$ cm, then the potential at the centre. Finally, say what changes if the sphere is a conductor.

The insulating sphere of radius

R=10.0

cm, charged throughout its volume, and the two points where the field is sought:

P_{1}

4.00

cm from the centre, inside, and

P_{2}

25.0

cm, outside. Distances to scale.

Step 1

Spherical symmetry: $\vec{E}$ is radial and its magnitude depends only on $r$ . Gaussian surface: a concentric sphere of radius $r$ .

Why

This is the line worth two marks out of five. Without it, you cannot write $\oint\vec{E}\cdot d\vec{A}=E(4\pi r^{2})$ , since pulling $E$ out of the integral assumes precisely that it is constant on the surface.

Step 2

For $r<R$ : $Q_{\text{enc}}=Q\dfrac{r^{3}}{R^{3}}$ , so $E=\dfrac{kQr}{R^{3}}$ . At $r=4.00$ cm: $E=1.62\times 10^{4}$ V/m.

Why

The word INSULATING in the problem is decisive: in a conductor, the charge would flee to the surface and there would be no enclosed charge. The field GROWS with $r$ because the charge grows as $r^{3}$ while the area only grows as $r^{2}$ .

Step 3

At $r=R$ , from the inside: $\dfrac{kQR}{R^{3}}=\dfrac{kQ}{R^{2}}=4.04\times 10^{4}$ V/m. From the outside: $\dfrac{kQ}{R^{2}}$ , identical.

Why

The matching is the mandatory check. If the two branches do not meet, there is an error, and finding it here costs thirty seconds against several minutes at the end of a test.

Step 4

For $r>R$ : $Q_{\text{enc}}=Q$ , so $E=\dfrac{kQ}{r^{2}}$ . At $r=25.0$ cm: $E=6.47\times 10^{3}$ V/m.

Why

Outside, everything happens as if the charge were a point charge at the centre: this is true for any distribution with spherical symmetry, hollow, solid or of varying density.

Step 5

Potential: $V(r)=\dfrac{kQ}{2R^{3}}(3R^{2}-r^{2})$ for $r\le R$ , so $V(0)=\dfrac{3}{2}\dfrac{kQ}{R}=6.07\times 10^{3}$ V, against $4.04\times 10^{3}$ V at the surface.

Why

The potential is MAXIMAL at the centre while the field is zero there, and that is consistent: $E=-\dfrac{dV}{dr}$ , and a function at its maximum has a zero derivative. It is the best example in the chapter that the value of $V$ says nothing about that of $E$ .

Step 6

CONDUCTING sphere: $E=0$ for every $r<R$ , $E$ unchanged for $r>R$ , and $V=4.04\times 10^{3}$ V constant throughout the volume.

Why

The outside field absolutely cannot tell the two cases apart: that is the limit of the law. What changes, and changes completely, is the inside, where the charge has moved to the surface.

The conclusion, written out

“Insulating: $1.62\times 10^{4}$ V/m at $4.00$ cm, $4.04\times 10^{4}$ V/m at the surface, $6.47\times 10^{3}$ V/m at $25.0$ cm, and $6.07$ kV at the centre. Conducting: zero field inside, uniform potential of $4.04$ kV, identical outside.”

The classic mistake on this problem: Applying the outside formula inside, or the reverse, without ever checking the matching. The other classic error is to forget the word insulating and conclude too quickly that the interior field is zero.

Learn by heart

• $\displaystyle\oint\vec{E}\cdot d\vec{A}=\dfrac{Q_{\text{enc}}}{\varepsilon_{0}}$ : only the ENCLOSED charge counts, and $\varepsilon_{0}=8.854\times 10^{-12}$ F/m.
• Three symmetries and only three: spherical, infinite cylindrical, infinite planar. Elsewhere, the law is true but unusable.
• Uniform ball: $E=\dfrac{kQr}{R^{3}}$ inside, $\dfrac{kQ}{r^{2}}$ outside. Wire: $\dfrac{2k\lambda}{r}$ . Sheet: $\dfrac{\sigma}{2\varepsilon_{0}}$ . Conductor: $\dfrac{\sigma}{\varepsilon_{0}}$ .
• A zero flux does not mean a zero field; a field that is zero everywhere on the surface, on the other hand, does give a zero flux.
• Non-uniform density: integrate $Q_{\text{enc}}=\displaystyle\int_{0}^{r}\rho(r')4\pi r'^{2}dr'$ BEFORE applying the law.
• Off-centre cavity: uniform field $\dfrac{\rho\vec{d}}{3\varepsilon_{0}}$ , independent of both radii.
• Potential of a uniform ball: $\dfrac{3}{2}\dfrac{kQ}{R}$ at the centre, one and a half times that of the surface.

Frequently asked questions

Does a zero flux mean that the electric field is zero?

No. Flux is a balance between what enters and what leaves the closed surface: a cube placed in a uniform field has a zero flux although the field has the same value at each of its points. A zero flux only indicates that the enclosed charge is zero. The converse is true: a field that is zero everywhere on the surface does give a zero flux.

When does Gauss's law really let you calculate the field?

Only when a symmetry provides a surface on which the magnitude of the field is constant and its direction known: spherical, infinite cylindrical or infinite planar symmetry. The law remains true everywhere else, but you can no longer pull the field out of the integral, and you must then go through superposition or a direct calculation.

Do charges outside the Gaussian surface create a field on it?

Yes, and it can be very intense: a charge of 50 nanocoulombs one centimetre away produces more than four million volts per metre. What is zero is their contribution to the net flux, because their field lines enter on one side of the closed surface and leave on the other. This is why only the enclosed charge appears in the law.

How does the field vary inside a uniformly charged insulating sphere?

It grows linearly with the radius, so it decreases as you approach the centre, where it vanishes. The charge enclosed by the Gaussian surface grows as the cube of the radius while the area of that surface only grows as the square: the quotient therefore grows as the radius.

Why is the field at the surface of a conductor double that of a charged sheet?

Because the Gaussian pillbox is not placed in the same way. For an insulating sheet, it extends on both sides and the flux leaves through its two end caps, hence the factor two in the denominator. For a conductor, one end cap is in the metal where the field is zero: all the flux leaves through a single end cap, and the factor two disappears.

How do you calculate the field when the charge density is not uniform?

You must first calculate the enclosed charge by integrating the density over the volume of the sphere of the radius considered, and only then apply the law. No formula learned for a uniform density applies: if the density grows as the radius, the enclosed charge grows as its fourth power and the field as its square, not as the radius.

Practise it

Corrected exercises: Gauss's law, 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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Electric charge, field and potential Next sheet Electric potential and capacitors

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

Looking for a 203-SN2-RE tutor in Montreal?

Get in touch for a first session. We work through Gauss's law at the level of real tests, from the choice of the surface to non-uniform densities.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: Gauss's law (203-SN2-RE)

The essentials

Flux: what passes through a surface

The law, and the only three symmetries that make it useful

What the law cannot do

Hollow conductor: the charges that appear

The mistakes that cost marks

1. A cube is placed in a uniform electric field of 350350350 N/C. The total flux through its surface is zero.

2. A charge of 505050 nC is placed 1.001.001.00 cm outside one face of the cube.

3. An insulating sphere of radius RRR carries a charge density ρ(r)=ρ0rR\rho(r)=\rho_{0}\dfrac{r}{R}ρ(r)=ρ0​Rr​, increasing outward. Calculate the field at r=R2r=\dfrac{R}{2}r=2R​.

4. A conducting plate carries σ=8.00\sigma=8.00σ=8.00 nC/m2^{2}2 on each of its faces. Calculate the field just outside.

5. Compare the outside field of a hollow sphere and a solid sphere carrying the same charge.

6. You are asked for the field in a spherical cavity hollowed out away from the centre of a uniformly charged ball.

Which method to choose

Which Gaussian surface, according to the symmetry

How the answer is expected to be written

Applying Gauss's law properly

Check before you hand in

The typical problem, taken apart

The solid insulating sphere, its two regions and their matching

Learn by heart

Frequently asked questions

Does a zero flux mean that the electric field is zero?

When does Gauss's law really let you calculate the field?

Do charges outside the Gaussian surface create a field on it?

How does the field vary inside a uniformly charged insulating sphere?

Why is the field at the surface of a conductor double that of a charged sheet?

How do you calculate the field when the charge density is not uniform?

Corrected exercises: Gauss's law, 203-SN2-RE

See also

Looking for a 203-SN2-RE tutor in Montreal?

1. A cube is placed in a uniform electric field of $350$ N/C. The total flux through its surface is zero.

2. A charge of $50$ nC is placed $1.00$ cm outside one face of the cube.

3. An insulating sphere of radius $R$ carries a charge density $\rho(r)=\rho_{0}\dfrac{r}{R}$ , increasing outward. Calculate the field at $r=\dfrac{R}{2}$ .

4. A conducting plate carries $\sigma=8.00$ nC/m $^{2}$ on each of its faces. Calculate the field just outside.