Review/ Practice Problems before Exam 2

In addition to this page, I will include the “purple equation sheet” from Griffiths.

No phones or other device that connects to the internet.

You may use a calculator, though I don’t think you’ll need it.

Present clear and complete answers:

Explain your answers clearly but briefly. You want to aim for a level of solution that someonetaking this class would be able to understand. A diagram and a few words may help.

Start calculations with first principles: things like definitions ( ~E ≡ ~FQ

) or empirical laws (like

Coulomb’s Law or Newton’s Laws) or conservation laws.

Check time:

The point values for each problem are shown next to the question number. Time yourselfaccordingly. The total value of the exam is 100 points. Good luck!

Some definitions:

8 Chapter 1 Vector Analysis

(ii) Vector triple product: A × (B × C). The vector triple product can besimplified by the so-called BAC-CAB rule:

A × (B × C) = B(A · C) − C(A · B). (1.17)

Notice that

(A × B) × C = −C × (A × B) = −A(B · C) + B(A · C)

is an entirely different vector (cross-products are not associative). All higher vec-tor products can be similarly reduced, often by repeated application of Eq. 1.17,so it is never necessary for an expression to contain more than one cross productin any term. For instance,

(A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C);A × [B × (C × D)] = B[A · (C × D)] − (A · B)(C × D). (1.18)

Problem 1.5 Prove the BAC-CAB rule by writing out both sides in componentform.

Problem 1.6 Prove that

[A × (B × C)] + [B × (C × A)] + [C × (A × B)] = 0.

Under what conditions does A × (B × C) = (A × B) × C?

1.1.4 Position, Displacement, and Separation Vectors

The location of a point in three dimensions can be described by listing itsCartesian coordinates (x, y, z). The vector to that point from the origin (O)is called the position vector (Fig. 1.13):

r ≡ x x̂+ y ŷ+ z ẑ. (1.19)

r

y

z

z

yx

x

(x, y, z)r

O

FIGURE 1.13

r

rr!

Source point

Field pointO

FIGURE 1.14

Some more math:

r 2 = r2 + r′2 − 2rr′ cosα

Vs =4

3πR3

Helpful Equations:

~F =qQ

4π�0 r 2r̂∮

~E · d~a = qenc�0

V (r) = −∫ rref

~E · d~̀

Helpful Integrals:∫ √1− x2dx = 1

2[x√

1− x2 + sin−1 x]

∫ dx√1− x2 = sin

−1 x

∫ xdx√1− x2 = −

√1− x2

∫ x2dx√1− x2 = −

x

2

√1− x2 + 1

2sin−1 x

Helpful Taylor series expansions (forsmall �):

e� ≈ 1 + �+ . . .ln(1 + �) ≈ �+ . . .

(1 + �)n ≈ 1 + n�+ . . .

88 Chapter 2 Electrostatics

Problem 2.28 Use Eq. 2.29 to calculate the potential inside a uniformly chargedsolid sphere of radius R and total charge q. Compare your answer to Prob. 2.21.

Problem 2.29 Check that Eq. 2.29 satisfies Poisson’s equation, by applying theLaplacian and using Eq. 1.102.

2.3.5 Boundary Conditions

In the typical electrostatic problem you are given a source charge distributionρ, and you want to find the electric field E it produces. Unless the symmetryof the problem allows a solution by Gauss’s law, it is generally to your advan-tage to calculate the potential first, as an intermediate step. These are the threefundamental quantities of electrostatics: ρ, E, and V . We have, in the courseof our discussion, derived all six formulas interrelating them. These equationsare neatly summarized in Fig. 2.35. We began with just two experimental obser-vations: (1) the principle of superposition—a broad general rule applying to allelectromagnetic forces, and (2) Coulomb’s law—the fundamental law of electro-statics. From these, all else followed.

You may have noticed, in studying Exs. 2.5 and 2.6, or working problems suchas 2.7, 2.11, and 2.16, that the electric field always undergoes a discontinuitywhen you cross a surface charge σ . In fact, it is a simple matter to find the amountby which E changes at such a boundary. Suppose we draw a wafer-thin Gaussianpillbox, extending just barely over the edge in each direction (Fig. 2.36). Gauss’slaw says that

∮

S

E · da = 1ϵ0

Qenc =1ϵ0

σ A,

where A is the area of the pillbox lid. (If σ varies from point to point or the surfaceis curved, we must pick A to be extremely small.) Now, the sides of the pillbox

ρ

EV

E =

ρdτ

14πε0

r 2 r

∇ .E =

ρ/ε0 ;∇

× E

= 0

V = − E.dl

V =

dτ

14π

ε 0rρ

∇2 V

= −ρ

/ε 0

E = −∇V

FIGURE 2.35

250 Chapter 5 Magnetostatics

J

BA

B =

dτ

µ04π

r 2J r

B = µ0 J; B = 0

A =

dτ

µ 04π

rJ2 A

= −

µ 0J

B = A; A = 0∆

∆

∆

∆

∆

×

×

×.

.

FIGURE 5.48

Just as the electric field suffers a discontinuity at a surface charge, so the mag-netic field is discontinuous at a surface current. Only this time it is the tangentialcomponent that changes. For if we apply Eq. 5.50, in integral form,

∮B · d a = 0,

to a wafer-thin pillbox straddling the surface (Fig. 5.49), we get

B⊥above = B⊥below. (5.74)

As for the tangential components, an Amperian loop running perpendicular to thecurrent (Fig. 5.50) yields

∮B · d l =

(B∥above − B∥below

)l = µ0 Ienc = µ0 Kl,

or

B∥above − B∥below = µ0 K . (5.75)

B⊥above

B⊥below

K

FIGURE 5.49

W =�02

∫E2dτ

~F = Q~E +Q~v × ~B ~F =∫Id~̀× ~B

~B(~r) =µ0I

4π

∫ d~̀× r̂r 2

∮~B · d~̀= µ0Ienc

Phys 110, Fall 2018, Exam 2 review 1

1. Remember the triangle we filled in (maybe even for the Exam 1 review?), shown here.Do it for the square! The square is mildly less interesting, but it shows the varioussymmetries between the quantities/equations nicely.

88 Chapter 2 Electrostatics

Problem 2.28 Use Eq. 2.29 to calculate the potential inside a uniformly chargedsolid sphere of radius R and total charge q. Compare your answer to Prob. 2.21.

Problem 2.29 Check that Eq. 2.29 satisfies Poisson’s equation, by applying theLaplacian and using Eq. 1.102.

2.3.5 Boundary Conditions

In the typical electrostatic problem you are given a source charge distributionρ, and you want to find the electric field E it produces. Unless the symmetryof the problem allows a solution by Gauss’s law, it is generally to your advan-tage to calculate the potential first, as an intermediate step. These are the threefundamental quantities of electrostatics: ρ, E, and V . We have, in the courseof our discussion, derived all six formulas interrelating them. These equationsare neatly summarized in Fig. 2.35. We began with just two experimental obser-vations: (1) the principle of superposition—a broad general rule applying to allelectromagnetic forces, and (2) Coulomb’s law—the fundamental law of electro-statics. From these, all else followed.

You may have noticed, in studying Exs. 2.5 and 2.6, or working problems suchas 2.7, 2.11, and 2.16, that the electric field always undergoes a discontinuitywhen you cross a surface charge σ . In fact, it is a simple matter to find the amountby which E changes at such a boundary. Suppose we draw a wafer-thin Gaussianpillbox, extending just barely over the edge in each direction (Fig. 2.36). Gauss’slaw says that

∮

S

E · da = 1ϵ0

Qenc =1ϵ0

σ A,

where A is the area of the pillbox lid. (If σ varies from point to point or the surfaceis curved, we must pick A to be extremely small.) Now, the sides of the pillbox

ρ

EV

E =

ρdτ

14πε0

r 2 r

∇ .E =

ρ/ε0 ;∇

× E

= 0

V = − E.dl

V =

dτ

14π

ε 0rρ

∇2 V

= −ρ

/ε 0

E = −∇V

FIGURE 2.35

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2. A parallel plate capacitor is made of two plates of surface area, A, and they are adistance, d apart. They each have a uniformly distributed equal charge: one positiveand one negative. Recall the the electric field near one plate of charge with chargedensity σ is given by:

~E =σ

2�0n̂

where n̂ means away from, and normal to, the plane of the plate.

(a) Find V between the plates in terms of σ, A, d and any physical constants.

(b) Find the capacitance of the plates in terms of σ, A, d and any physical constants.

(c) How much work does it take to move a new charge Q from the negative to thepositive plate? (New means, I’m not taking charge from one plate and moving itto the next. The charge on the plates stays the same. Just move a new chargefrom one to the other.)

(d) If you doubled the charge on each plate, how would the capacitance change? Howwould the work change?

Phys 110, Fall 2018, Exam 2 review 2

3. A very crude model for a neutral water molecule is shown in the figure, a negative “O”and two positive “H’s.” The geometry has been simplified to make the math easier. Asusual, assume V (∞) = 0.How much work does it take to configure this molecule, bringing each of the pieces(the O and two H’s) in from infinity and putting them in place?

4.) A very crude model for a neutral water molecule is shown in the figure, a negative “O” and two positive “H’s”. The geometry has been simplified to make the math easier. As usual, assume V(∞)=0. a) (4 pts) What is the Electric field at the origin (the point midway between the two positive hydrogens)? b) (2 pts) What is the electric potential at the origin? b-ii) Could you have integrated over the E-field to get V ? Why or why not? c) (4 pts) If you place a positive test charge (q’) at the origin, what is the force on the test charge (q’)?

x

y

a a

a +q/2 +q/2

-q

4. A 4m x 8m loop of wire carries a current of 5A. It is in a region with a 2T magneticfield. Calculate the force (magnitude and direction) on just the left segment.

5. A loop of wire (radius R) sits in the x-z plane (the plane of this page), with steadycurrent I running counter-clockwise around it, as shown. It is centered on the origin.

Now consider a dashed “Amperian loop” neatly surrounding part of the wire (it alsohas radius R, and lies in the x-y plane, so the figure shows a perspective view. Thisimaginary Amperian loop runs in and out of the plane of the paper)

(a) Is Ampere’s law TRUE for this loop? Why or why not?

(b) Can we USE Ampere’s law to simply compute the B-field at the origin? If yes,do it. If no, explain why not.

p. 5

Q3.) A loop of wire (radius R) sits in the x-z plane (the plane of this page), with steady current I running counter-clockwise around it, as shown. It is centered on the origin. Now consider a dashed “Amperian loop” neatly surrounding part of the wire (it also has radius R, and lies in the x-y plane, so the figure shows a “perspective view. This imaginary Amperian loop runs in and out of the plane of the paper) a) Is Ampere’s law TRUE for this loop? (2 pts) Circle one: Yes No b) Can we USE Ampere’s law to simply compute the B-field at the origin? (4 pts) Circle one: Yes No If yes, do it. If no, tell me briefly but clearly why not.

I

x

z

Phys 110, Fall 2018, Exam 2 review 3

6. You have an infinitely tall, infinitesimally thin cylindrical sheet of surface currentrunning around the z-axis,

~K =

Kφ̂ for s = R0 for s 6= RFind ~B inside the cylinder.

p. 3

Q1.) You have an infinitely tall, infinitesimally thin cylindrical sheet of surface current running around the z-axis, i.e. K =K0ϕ̂ at s=R (otherwise 0) . (You do not have to explain your work on this problem for full credit, although it might help you get partial credit if you make a simple mistake.) i) Sketch the magnitude of the resulting B field as a function of radial distance, s (both inside and outside the sheet) below. (4 pts) Please label the “tickmark” on the vertical graph with given quantities so I know the scale.

s

|B|(s)

R

ii) In what direction does the B field point at the center? (Or, if it is zero there, say so) (2 pts)

K

s

z

R

7. Two infinitely long current carrying wires are arranged as shown in the diagram below,the top one with current 6µA going into the page, the other with current 6µA comingout of the page. a = 30 cm and b = 40cm. What is the magnetic field, B, at point P?

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