physics quiz 2

8/13/2019 Physics Quiz 2

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Physics 2212 Quiz #2 SolutionsFall 2011

0 Permittivity constant mp Mass of a protone Fundamental charge me Mass of an electron

K Coulomb constant = 1/40 g Gravitational acceleration

Unless otherwise directed, friction, drag, and gravity should be neglected.

Unless otherwise directed, any integrals in free-response problems must be evaluated.

I. (16 points) A circular disk with a hole in its middle has an inner radius,a, and an outer radius, b, and hasa chargeQ uniformly distributed on its surface. Letr be the distance from the center of the disk to a pointon the disk. Set up an integral for the electric potential on the axis of the disk at a distancex from thecenter, in terms of parameters defined in the problem and physical or mathematical constants, but do notevaluate the integral!

. . . . . . . . . . . . . . .

Choose an element of charge, dQ, that is all the same distance fromthe point x where the electric potential is to be found. This elementwill be a ring of radius r and width dr . The charge on the ring is thearea charge density of the disk times the the area of the ring. The areaof the ring is its circumference times its width.

dQ= dA = 2r dr

Since every point on this ring is the same distance r =

x2 +r2 fromthe pointx, the potential due to the ring is the same as would be dueto a point charge at that same distance.

dV =KdQ

r =K

2r drx2 +r2

Add up all the rings dQ, which is an integral from a to b. Note that since the charge on the disk is uniformlydistributed, is a constant, and is the total charge on the disk qdivided by the area of the disk b2 a2.

V =

dV =

ba

K2r dr

x2 +r2=K 2

ba

r drx2 +r2

= KQ2

b2 a2 ba

r drx2 +r2

= 2KQ

b2 a2 ba

r drx2 +r2

Quiz #2 Solutions Page 1 of 7


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II. (16 points) An infinitely long solid insulating cylinder of radius R has a volume charge density that varieswith the radius as

= 0

a r

b

Where 0

, a, and b are positive constants and r is the distance from the axis of the cylinder. Find anexpression for the magnitude of the electric field in the region r < R, in terms of parameters defined in theproblem and physical or mathematical constants.

. . . . . . . . . . . . . . . . . . . . . . .Use Gauss Law. Choose a Gaussian Surface with thesymmetry of the charge distribution, and that passesthrough the point at which the field is to be deter-mined. The sketch shows such a Gaussian Surface, acylinder with radius r and length. The flux throughthis surface is proportional to the charge within it:

0 = qin 0

E dA=

dV

By choosing the Gaussian Surface wisely, the flux integral has been made straightforward. The flux is zerothrough the ends of the surface, as the field (which must be directed straight away from the central axis dueto symmetry considerations) is perpendicular to the area vector there. On the curved part of the surface,the electric field has constant magnitude, and is parallel to the area vector at each point. The area of thecurved part is the circumference of the Gaussian Surface times its length.

=

E dA= E

curved

dA= EAcurved= E2r

Since the charge density varies with r, when finding the charge within the Gaussian Surface, the volumeelement dV should be chosen to be small in the r direction. A thin cylindrical shell has been chosen, withradius r, thickness dr, and length . The volume of this shell is its area times its thickness. Note that at

a distancer from the axis, the charge density is 0(a r/b). The charge in all the thin shells nested fromradius zero to radius r must be added up.

qin =

dV =

r0

0

a r

b

2r dr= 20

r0

ar r

2

b

dr

= 20

ar2

2 r

3

3b

r0

= 20

ar2

2 r

3

3b

Now that both the flux and charge have been determined, solve for the field magnitude.

0 = qin 0E2r = 20

ar2

2 r

3

3b

E= 0

0

ar

2 r

2

3b

1. (6 points) How will the electric field magnitude vary as a function of the distance,r, whenr is large comparedto the cylinder radius R (i.e.,r >> R)?

. . . . . . . . . . . . . . . . . . . . . . .

When r is large compared to the cylinder radius R, the field must decrease like that of a uniform line ofcharge. That is, it must decrease faster than the field due to a uniform infinite sheet (which is constant),but not as fast as the field due to a point charge (which decreases with 1/r2).

E 1/r



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III. (16 points) Two positive charges Q are fixed at the vertices of anequilateral triangle with sides of length a. A particle of positivecharge, q, and mass m is positioned at the apex of the equilateraltriangle as shown. It is released from that point with an initial

velocity ofv 0along the center line shown. What must the minimumspeed of this particle be so that it passes the two fixed charges?Express your answer in terms of parameters defined in the problemand physical or mathematical constants.

. . . . . . . . . . . . . .

Use the Work-Energy Theorem,Wext+Wnc = K+ U

Choose a system consisting of all three charged particles. There is no work done by external forces on thatsystem, and there is no work done by non-conservative forces within that system. The potential energy isthat of a system of charged particles.

0 + 0 = (Kf Ki) + (Uf Ui)

So

0 =12

mv2f 12mv2i

+

K

Qq

rfKQq

ri

+

K

Qq

rfKQq

ri

+

K

Q2

rf KQ

2

ri

where the potential energy of the system is the sum of the potential energies associated with each pair ofparticles. Note that the final speed of the particle with chargeqmust be only infinitesimally more than zeroas it passes between the particles with charge Q.

0 =

0 12

mv20

+

K

Qq

a/2 KQq

a

+

K

Qq

a/2 KQq

a

+

K

Q2

a KQ

2

a

Solve forv0:

12mv20 = 2

KQq

a/2 KQq

a

+ 0 = 2KQq

a v0 =4K

m

Qq

a

2. (6 points) Consider a situation in which the particle with chargeqin the problem above were replaced by aparticle with charge q = 2q, and the fixed charges Q were each replaced with fixed charges Q = 2Q. Howdoes the minimum speed, v

0, required for the particle to pass the fixed charges in this situation, compare to

your answer v0 above?

. . . . . . . . . . . . . . . . . . . . . . .

Since the electric potential energy depends on the product of the pairs of charges, doubling all the chargevalues will increase the potential energy by a factor of 4. Thus, the moving particle will need four times thekinetic energy to pass between the fixed charges. As the kinetic energy depends on the square of the speed,doubling the speed will provide four times the kinetic energy.

v0

= 2v0



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3. (5 points) A hollow conductor ispositivelycharged. A small uncharged metal ball is lowered by a silk threadthrough a smallopening in the top of the conductor and allowed to touch its inner surface. After the ball isremoved, it will have . . .

. . . . . . . . . . . . . . . . . . . . . . .

Since the hole is small, we can treat the inner and outer surfaces of the hollow conductor as if they wereseparate. Gauss Law can be used to show that the excess charge on the hollow conductor must reside onits outer surface. The uncharged metal ball only touches the uncharged inner surface, and so it gains . . .

no appreciable charge.

4. (5 points) A Gaussian surface completely encloses an electric dipole of dipole moment magnitude p andcharge separation magnitude d. The total electric flux due to the dipole through this Gaussian surface

. . . . . . . . . . . . . . . . . . . . . . .An electric dipole consists of equal and opposite charges separated by some distance. As this dipole liesentirely within the Gaussian Surface, there is no net charge, then, within the Gaussian Surface. Gauss Lawstates that the net flux through a closed surface is proportional to the net charge within it. As the net chargeis zero, the total electric flux . . .

is zero.



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5. (5 points) A long line of charge with linear charge densityL runs along the axis of a conducting cylindricalshell which carries a charge per unit length ofC. The charge per unit length on the inner and outer surfacesof the shell are . . .

. . . . . . . . . . . . . . . . . . . . . . .

Consider a cylindrical Gaussian Surface whose curved surface lies within the thickness of the conductingcylindrical shell. There is no net flux through this surface, as any field between the line of charge and theinner surface of the conductor is parallel to the area vectors, and there is no field within the thickness ofthe conductor. Since Gauss Law states that the net flux through a closed surface is proportional to thenet charge within it, must be a net charge of zero within the surface. Thus, there must be a charge densityL on the inner surface to cancel the charge density L of the line. As charge is conserved, however, thenet charge and charge density on the cylinder cannot change. If the net charge density remainsC, thenC+ L must reside on the outer surface.

L on the inner surface, andC+ L on the outer surface.

6. (5 points) An electron moves from pointi to pointf, in the direction of a uniform electric field. During thisdisplacement. . .

. . . . . . . . . . . . . . . . . . . . . . .

As an electron is negative (q=e), the force of the field on the electron, F =qE=e E, is opposite thedirection of the field. Therefore, the force of the field on the electron is opposite the displacement of the

electron, and the work done by the field on the electron is negative. Since this work is done by a force internalto the field-electron system, the change in potential energy is of the opposite sign (that is, positive). Recallthat the relationship between work done by internal forces and potential energy changes is Wint= U.

(By analogy, when the distance between an object and the Earth increases, the gravitational force does

negative work on the object, and the potential energy of the Earth-object system increases.) Sothe work done by the field is negative and the potential energy of the electron-field system increases.



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7. (5 points) The figure below shows four arrangements of charged particles, all the same distance from theorigin. If possible, rank the situations according to the electric potential at the origin, from greatest (mostpositive) to least.

. . . . . . . . . . . . . . . . . . . . . . .

Since all the particles are the same distance from the origin, the potential at the origin is the same as itwould be if there were a single particle with the total charge of the arrangement. The issue of the zeropoint is irrelevant, as redefining the zero point would have the same effect on all four arrangements. The netcharge in each arrangement is:

Qi = 7q,

Qii = 5q,

Qiii = 11q,

Qiv = 8q, so

ii > i > iv > iii

Note that there is a single exception to the zero point having the same effect on all four arrangements. Thatwould be if the origin itself were selected as the zero point. Full credit, therefore, was issued for

Ranking is impossible unless location of zero potential is specified.

8. (5 points) The diagram shows four pairs of large parallel conducting plates. The value of the electric potentialis given for each plate. Rank the pairs according to the magnitude of the electric field between the plates,greatest to least.

. . . . . . . . . . . . . . . . . . . . . . .

AsEs = V/s, the magnitude of the uniform field in these capacitors can be calculated from E= V /s,yieldingEi = 2V0/d, Eii = V0/2d, Eiii = V0/4d, and , Eiv = V0/d, so

i > iv > ii > iii



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9. (5 points) A positive point charge +q lies at the center of a tetrahedron, constructed of four equilateraltriangles with edges a. What is the electric flux through the bottom face of the tetrahedron?

. . . . . . . . . . . . . . . . . . . . . . .

Each face is identical, so one-fourth of the total flux must pass through each face.

Use Gauss Law.0 = qin /4 =+q/40

10. (5 points) A solid insulating sphere of radius R contains a uniform volume distribution of positive charge.Which of the following graphs below correctly gives the magnitudeEof the electric field as a function ofr ,the distance from the center?

. . . . . . . . . . . . . . . . . . . . . . .

The electric field at the center of the sphere must be zero due to symmetry considerations. The electric fieldoutside the sphere must vary with 1/r2, as if the charge were concentrated at a point at the center of thesphere. The field magnitude at the surface of the sphere must be the same whether it is approached frominside or outside the sphere. These considerations eliminate three of the five graphs proposed. Gauss Law

can be used (see example 28.4 of your text) to prove that the field magnitude inside the sphere is linearlyproportional to r .


physics quiz 2

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