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  -  2  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
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