Problem 1 solution

1 Part a

1.1 solution

The easiest way to calculate the potential is to take advantage of the eﬀective spherical symmetry of

inﬁnitesimal charge elements

−Zl

E·dl =−Zl

dl ·ZQ

kdq

r2ˆr

=−ZQ

kdq Zr

∞

dr ·ˆr

r2=ZQ

kdq

r

The apparent cylindrical symmetry of the problem encourages cylindrical coordinates, in which the charge

element is dq =σρdρdθ, and r=pρ2+a2

kZa

0

dρ Z2π

0

σρdρdθ

pρ2+z2=kZz2+a2

z2

πσdu

u1

2

= 2πσku 1

2|z2+a2

z2= 2πkσ ³pz2+a2−√z2´

=σ

2²k³pz2+a2−z´

The usubstitution being u=z2+a2=⇒du =zdz

1.2 rubric

3 points for recognizing cylindrical symmetry.

3 points for distinguishing rfrom ρ(!!), writing rcorrectly, writing dq correctly, and setting up the integral

2 for calculating correctly and reporting the potential as a scalar

2 part b

E=−∇V. Since there is only zdependence

E=−∂V

∂z ˆz

=σ

2²µ1−z

√z2+a2¶ˆz

2.1 rubric

4 points for setting up the calculation correctly

3 points for recognizing the potential only has dependence z. (Some thought there was adependence, but

aparametrizes the radius of the disk, and is not a coordinate).

1 point for including vector direction of E

3 part c

By superposition,

Eplanewithaholeinit =Eplane −Edisk

=σ

2²ˆz −σ

2²µ1−z

√z2+a2¶ˆz

=σ

2²

z

√z2+a2

ˆz

3.1 rubric

5 points for seeing superposition

2 points for computing properly

1 point for remembering the ˆz

7B MT2 Huang Solutions Patrick Varilly

1 Problem 2

(a)

[8pts for (a)] In terms of Rand R0, the resistance from A to B can be decomposed as follows:

Series[ R0, Parallel[ R0, Series[ R,R0] ] ].

Thus, the eﬀective resistance from A to B is given by

Rtot =R0+1

1

R0+1

R+R0

.

[Up to 6pts for this expression, partial credit for a partially correct answer] We can manip-

ulate this expression to obtain

Rtot =R0+R+R0+R0

R0(R+R0)−1

,

=R0+R0(R+R0)

R+ 2R0,

=R0(R+ 2R0+R+R0)

R+ 2R0,

=2R+ 3R0

R+ 2R0·R0.

Setting Rtot =Rand solving for R0yields:

R=2R+ 3R0

R+ 2R0·R0,

R(R+ 2R0) = (2R+ 3R0)R0,

R2+ 2R0R= 2RR0+ 3(R0)2,

(R0)2=R2/3.

Hence

R0=1

√3R .

[2 pts for solving for R0correctly]

(b)

[12 pts for all of (b). This part can be solved in many diﬀerent ways: the grading scheme

reﬂects one particularly straightforward approach, but it is possible to get full credit if you

1

7B MT2 Huang Solutions Patrick Varilly

solve the problem in a diﬀerent manner. One complication is that the manipulations with

square roots of 3 can be quite hairy: I’ve thus correspondingly dropped very few points for

mistaken algebra, emphasizing the points on the physical concepts: Voltages across parallel

branches are equal, and they distribute over a sequence of elements in series; current divides

at a junction; Ohm’s law applies to all the resistors; etc.]

We ﬁrst calculate the answers in terms of R0and V, then use the result from (a) to

express these in terms of Vand Ronly. As a check, we know that the total power dissipated

should be

Ptot =V2

Rtot

=V2

R.

Since Rtot =R, we know that the total current ﬂowing from Ato Bmust be given by

[1pt]

Itot =V

R.

Denote by Ii,Viand Pithe current ﬂowing through resistor i, the voltage drop across it,

and the power dissipated through it. Since I4=Itot, we have

V4=ItotR0=VR0

R,

We then have [2pt]

P4=V2

4/R0=V2R0

R2.

The voltage across the rest of the circuit is

Vrest =V−V4=VR−R0

R.

Since the rest of the circuit consists of resistors 1 and 2 and resistor 3 in parallel, we have

[2pt]

V3=V1+2 =Vrest.

Thus [1pt],

P3=V2

3/R0=V2(R−R0)2

R2R0.

To examine resistors 1 and 2, we note that I1=I2and V1+V2=V1+2, so

I1(R1+R3) = Vrest =VR−R0

R,

so [2pt]

I1=I2=VR−R0

R(R+R0).

2

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University:
Assam Agricultural University

Subject:
Engineering Physics

Upload date:
12/02/2013