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

##### Document information

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sashie

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Address:
Physics

University:
Assam Agricultural University

Subject:
Engineering Physics

Upload date:
12/02/2013