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

Engineering Physics

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 important points are: Infinitesimal Charge Elements, Effective Spherical Symmetry, Cylindrical Coordinate, Effective Resistance, Ohm Law, Total Power, Vector Direction
Showing pages  1  -  4  of  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
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
Z2π
0
σρdρdθ
pρ2+z2=kZz2+a2
z2
πσdu
u1
2
= 2πσku 1
2|z2+a2
z2= 2πkσ ³pz2+a2z2´
=σ
2²k³pz2+a2z´
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
2²µ1z
z2+a2ˆz
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
2²µ1z
z2+a2ˆz
=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 =VV4=VRR0
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(RR0)2
R2R0.
To examine resistors 1 and 2, we note that I1=I2and V1+V2=V1+2, so
I1(R1+R3) = Vrest =VRR0
R,
so [2pt]
I1=I2=VRR0
R(R+R0).
2