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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

*−
*∫

*l
*E *· *dl = *−
*

∫

*l
*dl *·
*

∫

*Q
*

*kdq
*

*r*2
r̂

= *−
*∫

*Q
kdq
*

∫ *r
∞
*

dr *· *r̂
*r*2

= ∫

*Q
*

*kdq
*

*r
*

The apparent cylindrical symmetry of the problem encourages cylindrical coordinates, in which the charge
element is *dq *= *σρdρdθ*, and *r *=

√
*ρ*2 + *a*2

*k
*

∫ *a
*0

*dρ
*

∫ 2*π
*0

*σρdρdθ*√
*ρ*2 + *z*2

= *k
*∫ *z*2+*a*2

*z*2

*πσdu
*

*u
*1
2

= 2*πσku
*1
2 *|z*2+*a*2

*z*2
= 2*πkσ
*

(√
*z*2 + *a*2 *−
*

*√
z*2

)

=
*σ
*

2*²
k
*

(√
*z*2 + *a*2 *− z
*

)

The *u *substitution being *u *= *z*2 + *a*2 =*⇒ du *= *zdz
*

1.2 rubric

3 points for recognizing cylindrical symmetry.
3 points for distinguishing *r *from *ρ*(!!), writing *r *correctly, 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 *z *dependence

E = *−∂V
∂z
*

ẑ

=
*σ
*

2*²
*

(
1*− z√
*

*z*2 + *a*2

) ẑ

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 *a *dependence, but
*a *parametrizes the radius of the disk, and is not a coordinate).
1 point for including vector direction of E

3 part c

By superposition,

E*planewithaholeinit *= E*plane −*E*disk
*=

*σ
*

2*²
*ẑ− *σ
*

2*²
*

(
1*− z√
*

*z*2 + *a*2

) ẑ

=
*σ
*

2*²
z√
*

*z*2 + *a*2
ẑ

3.1 rubric

5 points for seeing superposition 2 points for computing properly 1 point for remembering the ẑ

7B MT2 Huang Solutions Patrick Varilly

1 Problem 2

(a)

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

Series[ R′, Parallel[ R′, Series[ R, R′ ] ] ].

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

Rtot = R ′ +

1 1 R′

+ 1 R+R′

.

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

Rtot = R ′ +

( R +R′ +R′

R′(R +R′)

)−1 ,

= R′ + R′(R +R′)

R + 2R′ ,

= R′(R + 2R′ +R +R′)

R + 2R′ ,

= 2R + 3R′

R + 2R′ ·R′.

Setting Rtot = R and solving for R ′ yields:

R = 2R + 3R′

R + 2R′ ·R′,

R(R + 2R′) = (2R + 3R′)R′,

R2 + 2R′R = 2RR′ + 3(R′)2,

(R′)2 = R2/3.

Hence

R′ = 1√ 3 R .

[2 pts for solving for R′ correctly]

(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 R′ and V , then use the result from (a) to express these in terms of V and R only. As a check, we know that the total power dissipated should be

Ptot = V 2

Rtot = V 2

R .

Since Rtot = R, we know that the total current flowing from A to B must be given by [1pt]

Itot = V

R .

Denote by Ii, Vi and Pi the current flowing through resistor i, the voltage drop across it, and the power dissipated through it. Since I4 = Itot, we have

V4 = ItotR ′ = V

R′

R ,

We then have [2pt]

P4 = V 2 4 /R

′ = V 2 R′

R2 .

The voltage across the rest of the circuit is

Vrest = V − V4 = V R−R′

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 = V 2 3 /R

′ = V 2 (R−R′)2

R2R′ .

To examine resistors 1 and 2, we note that I1 = I2 and V1 + V2 = V1+2, so

I1(R1 +R3) = Vrest = V R−R′

R ,

so [2pt]

I1 = I2 = V R−R′

R(R +R′) .

2

7B MT2 Huang Solutions Patrick Varilly

Hence [1pt each],

P1 = I 2 1R = V

2 (R−R′)2

R(R +R′)2 ,

and

P2 = I 2 2R ′ = V 2

(R−R′)2R′

R2(R +R′)2 .

To express these more compactly, let

φ := R′

R =

1√ 3 .

Then

P1 = V 2

R

(1− φ)2

(1 + φ)2 ,

P2 = V 2

R

(1− φ)2φ (1 + φ)2

,

P3 = V 2

R

(1− φ)2

φ ,

P4 = V 2

R φ.

Simplifying these, we get [2 pts for correct final answers in terms of V and R only, 1 pt if any incorrect, 0 pts if all incorrect]

P1 =

( 7− 12√

3

) · V

2

R ≈ 0.072V 2/R,

P2 =

( 7√ 3 − 4

) · V

2

R ≈ 0.041V 2/R,

P3 =

( 4√ 3 − 2

) · V

2

R ≈ 0.309V 2/R,

P4 = 1√ 3 · V

2

R ≈ 0.577V 2/R.

We can see that, indeed, P1 + P2 + P3 + P4 = V 2/R.

3

Midterm 2 Huang Solution of problem3 (by Haoyu)

Solution 3

When two capacitors are connected to each other, the charge will redistribute. Before the redistrubution, charge is

*Q*1 = *C*1*V*1 (1)
*Q*2 = *C*2*V*2 (2)

after the redistribution, suppose the charge will be *Q′*1 and *Q
′
*2, corresponding

voltages are *V ′*1 and *V
′
*2 , we will have

*V ′*1 = *V
′
*2 = *V
*

*′ *(3)

and because of the ‘island’ effect, we have

*Q′*1 + *Q
′
*2 = *Q*1 + *Q*2 (4)

therefore (a)

*C*1*V
′ *+ *C*2*V ′ *= *C*1*V*1 + *C*2*V*2 (5)

solve for *V ′ *we get

*V ′*1 = *V
′
*2 = *V
*

*′ *=
*C*1*V*1 + *C*2*V*2

*C*1 + *C*2
(6)

(b) The charges are easy to get

*Q′*1 = *C*1*V
′ *= *C*1

*C*1*V*1 + *C*2*V*2
*C*1 + *C*2

(7)

*Q′*2 = *C*2*V
′ *= *C*2

*C*1*V*1 + *C*2*V*2
*C*1 + *C*2

(8)

(c) When we connect the two capacitors in opposite, we can replace *V*2 with
*−V*2, and treat them the same way we did above. Therefore

*V ′′ *=
*C*1*V*1 *− C*2*V*2

*C*1 + *C*2
(9)

Positive sign of *V ′′ *means it is in the same direction as *V*1, and vice versa.

(d) The charge on each capacitor is given by

*Q′′*1 = *C*1*V
′′ *= *C*1

*C*1*V*1 *− C*2*V*2
*C*1 + *C*2

(10)

*Q′′*2 = *C*2*V
′′ *= *C*2

*C*1*V*1 *− C*2*V*2
*C*1 + *C*2

(11)

1