Problem 2.49

Electrostatics

Dustin Cassell ([email protected])

12/7/2011

Problem 2.49

Imagine that new and extraordinarily precise measurements have revealedan error in Coulomb's law. The actual force of interaction between two pointcharges is found to be

F =1

4πε0

q1q2<2

(1 +<λ

)e−</λ

∧<

where λ is a new constant of nature (it has dimensions of length, obviously,and is a huge numbersay half the radius of the known universeso that thecorrection is small, which is why no one ever noticed the discrepancy before).You are charged with the task of reformulating electrostatics to accommo-date the new discovery. Assume the principle of superposition still holds.

a) What is the electric eld of charge distribution ρ (replacing Equation 2.8)?

Equation 2.8 states,

E(r) =1

4πε0

ˆv

ρ(r′)

<2

∧< dτ

′.

So in our problem we have,

E(r) =1

4πε0

ˆρ(r

′)

<2

(1 +<λ

)e−</λ

∧< dτ.

b) Does this electric eld admit a scalar potential? Explain briey how youreached your conclusion.

1

Here we can problem 1.52, where we found that for a eld to admit a scalarpotential its curl had to be zero everywhere. Since our electric eld is radiallysymmetric, naturally its curl will be zero.

c) Find the potential of a point charge qthe analog to Equation 2.26. Use∞ as your reference point.

Equation 2.26 states,

V (r) =1

4πε0

q

<.

Our job is to nd the analog to this problem, so we note equation 2.21 whichstates,

V (r) = −r

OE · dl.

For our problem O =∞ and dl=dr. Hence we have,

V (r) = −r

∞

q

4πε0

1

r2

(1 +

r

λ

)e−r/λdr

=q

4πε0

− r

∞

1

r2e−r/λdr −

r

∞

1

rλe−r/λdr

=

q

4πε0

Ei(− rλ

)λ

0− e−r/λ

r|∞r − Ei

(−λr

) 0

where Ei is the exponential integral, and Ei=0 as r > 0. Thus we are leftwith,

V (r) =q

4πε

e−r/λ

r.

d) For a point charge q at the origin, show that˛S

E · da+1

λ2

ˆV

V dτ =1

ε0q,

where S is the surface, V is the volume, of any sphere centered at q.

2

First we note since→E

points radially outward,¸S

→E· da =

¸SE da.

We also note, E = q4πε0

1R2

(1 + R

λ

)e−R/λ, da = 4πR2, and nally dτ =

r2sinθdθdφdr. Hence,

˛S

E da+1

λ2

ˆV

V dτ =4πR2q

4πε0

1

R2

(1 +

R

λ

)e−R/λ

+1

λ2

R

0

2π

0

π

0

q

4πε0

e−r/λ

rr2sinθdθdφdr

=q

ε0

(1 +

R

λ

)e−R/λ +

1

λ2

q

ε0

r

0r e−r/λdr

=q

ε0

(1 +

R

λ

)e−R/λ +

q

λ2ε0

[−rλe−r/λ − λ2e−r/λ

]|R0

=q

ε0

(1 +

R

λ

)e−R/λ +

q

ε0

[−Re−R/λ

λ− eR/λ + 1

]

=q

ε0

[(1 +

R

λ

)e−R/λ −

(R

λ+ 1

)e−R/λ + 1

]=

q

e0.

e) Show that this result generalizes:

˛S

E · da+1

λ2

ˆV

V dτ =1

ε0Qenc,

for any charge distribution. (This is the next best thing to Gauss's Law, inthe new electrostatics.)

With the old Gauss's law we know ΦE =¸ →

E· da = Qenc

ε0=→∇ · →

E,

which can be proved in general by imagining some arbitrary surface as shownbelow

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Here we imagine S as a sphere, but note for each da on S, there exists ada' on S', which will subtend the same solid angle Ω. This implies thatΦ′ =

¸dΦ′ =

¸dΦ = Φ. Our point in noting this is twofold, if there is to be

a Gauss's Law in the new universe, then the idea of ux through a solid anglebetter hold true. The second reason is the important point that the solidangle subtended through any imaginable surface in conventional geometry is

4π. This is great, so let us now nd→∇ · →

Ein generalized coordinates.

But rst recall,

E(r) =1

4πε0

ˆρ(r′)

<2

(1 +<λ

)e−</λ

∧< dτ.

Now we will create a general coordinate to prove our equation holds forany charge distribution. We call this generalized the vector coordinate as ξ.Thus,

→∇ · →

E=

1

4πε0

ˆρ(ξ′)

→∇

(ξ − ξ′

||ξ − ξ′||3

)(1 +||ξ − ξ′||

λ

)e−||ξ−ξ

′||/λd3ξ′

where→∇ ·

(ξ−ξ′||ξ−ξ′||3

)= 4πδ(ξ − ξ′). Continuing we have,

→∇ · →

E=

1

4πε0

ˆρ(ξ′)

→∇

(ξ − ξ′

||ξ − ξ′||3

)e−||ξ−ξ

′||/λd3λ′

+1

4πε0

ˆρ(ξ′)

→∇

(ξ − ξ′

||ξ − ξ′||3

)(||ξ − ξ′||

λ

)e−||ξ−ξ

′||/λd3ξ′

4

=1

4πε0

ˆρ(ξ′)

(4πδ3 (ξ′) e−||ξ−ξ

′|| − 1

λ

(ξ − ξ′

||ξ − ξ′||3

)ξ − ξ′

||ξ − ξ′||e−||ξ−ξ

′||)d3ξ′

+1

4πε0

ˆρ(ξ′)[4πδ3(ξ′)

(||ξ − ξ′||

λ

)e−||ξ−ξ

′||/λ

+1

λ

ξ − ξ′

||ξ − ξ′||3ξ − ξ′

||ξ − ξ′||e−||ξ−ξ

′|| − ξ − ξ′

||ξ − ξ′||3||ξ − ξ′||

λ2

ξ − ξ′

||ξ − ξ′||e−||ξ−ξ

′||λ]d3ξ′.

=1

4πε0

ˆρ(ξ′)4πδ3(ξ′)

(1 +||ξ − ξ′||

λ

)e−||ξ−ξ

′||/λd3ξ′

− 1

4πε0

ˆρ(ξ′)

1

||ξ − ξ′||· 1

λe−||ξ−ξ

′||/λd3ξ′

=ρ

ε0− 1

λ2· 1

4πε0

ˆρ(ξ′)

||ξ − ξ′||e−||ξ−ξ

′||/λd3ξ′

=ρ

ε0− 1

λ2

ˆV

V dτ.

This implies that¸SE · da+ 1

λ2

´VV dτ = Qenc

ε0.

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