# Infinitesimal Charge Elements - Physics for Scientist and Engineers - Solved Past Paper, Exams for Engineering Physics. Assam Agricultural University

PDF (273 KB)
8 pages
956Number of visits
Description
This is the Solved Past Paper of Physics for Scientist and Engineers which includes Ideal Gas of Diatomic Molecules, Avogadro Number, Boltzmann Constant, Universal Gas Constant, Thermal Linear Expansion etc. Key importan...
20points
this document
Preview3 pages / 8

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

r2 r̂

=

Q kdq

r ∞

dr · r2

= ∫

Q

kdq

r

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

ρ2 + a2

k

a 0

∫ 2π 0

σρdρdθρ2 + z2

= k z2+a2

z2

πσdu

u 1 2

= 2πσku 1 2 |z2+a2

z2 = 2πkσ

(√ z2 + a2

√ z2

)

= σ

2² k

(√ z2 + a2 − z

)

The u substitution being u = z2 + a2 =⇒ 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

= σ

2²

( 1− z√

z2 + a2

) 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,

Eplanewithaholeinit = Eplane −Edisk =

σ

2² ẑ− σ

2²

( 1− z√

z2 + a2

) ẑ

= σ

2² z√

z2 + a2 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

Q1 = C1V1 (1) Q2 = C2V2 (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 = Q1 + Q2 (4)

therefore (a)

C1V ′ + C2V ′ = C1V1 + C2V2 (5)

solve for V ′ we get

V ′1 = V ′ 2 = V

= C1V1 + C2V2

C1 + C2 (6)

(b) The charges are easy to get

Q′1 = C1V ′ = C1

C1V1 + C2V2 C1 + C2

(7)

Q′2 = C2V ′ = C2

C1V1 + C2V2 C1 + C2

(8)

(c) When we connect the two capacitors in opposite, we can replace V2 with −V2, and treat them the same way we did above. Therefore

V ′′ = C1V1 − C2V2

C1 + C2 (9)

Positive sign of V ′′ means it is in the same direction as V1, and vice versa.

(d) The charge on each capacitor is given by

Q′′1 = C1V ′′ = C1

C1V1 − C2V2 C1 + C2

(10)

Q′′2 = C2V ′′ = C2

C1V1 − C2V2 C1 + C2

(11)

1