Amy W.

asked • 02/16/17

Electric Potential

Part a is as follows:
A long cylindrical insulator has a uniform charge density of 1.89 μC/m3 and a radius of 5 cm. What is the electric field inside the insulator at a distance of 4 cm? Answer in units of N/C.
 
I answered part a correctly using Gauss's Law. But now, I'm stuck on part b:
How much work must you do to bring a q = 0.032 μC test charge from 11 cm to 4 cm? Answer in units of J.
 
I figured I would need to integrate from 11 to 4 cm for the electric field and then multiply by q, but I'm a little confused. Can I integrate over kQ/r^2 from 11 to 4cm, then multiply by q?

Arturo O.

 
 
W = q∫0.110.04 E·dr
 
E is radial, so E·dr = Edr
 
Are you working this as a multiregion problem?
 
Gauss' law:
 
In the region from r = 0 to r = 0.05 m (inside the cylinder),
 
E(r) (2πr) = ρ/ε0 (πr2)
 
E(r) =  ρr / 2ε0 
 
In the region r > 0.05 m (outside the cylinder),
 
E(r) (2πr) = ρ/ε0 π(0.05 m)2
 
W = q∫0.110.04 E(r)dr = q∫0.110.05 [ρ/ε0 π(0.05 m)2] dr + q∫0.050.04 [ ρr / 2ε0] dr
 
Note you have to break up the work integral into 2 regions, inside and outside the cylinder, because the field has different forms inside and outside the cylinder.  Let me know if you need further explanation.
 
 
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02/16/17

Arturo O.

I forgot to divide by 2πr in the first integral.  The corrected expression is:
 
W = q∫0.110.05 [ρ/2ε0r (0.05 m)2] dr + q∫0.050.04 [ ρr / 2ε0] dr
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02/16/17

Steven W.

tutor
I think that is right on.  Just one quick comment that doesn't change the result, and maybe Arturo accounted for it for that reason.  There is actually a negative sign in front of the integral in the general definition of electric potential (and thus work done to move the test charge) as a path integral of the electric field.
 
V = - ∫E • ds
 
Here, the electric field from the cylindrical insulator points radially outward, as Arturo said.  This is why only the radial direction has to be integrated (this is mainly a symmetry argument from Gauss' law).  But, since the electric field points outward, and the path of the integral is radially inward (from 11 cm to 4 cm), the dot product inside the integral also contributes a -1 (from the cos of the 180 degree angle between E and dr at every point) that cancels out the negative sign from the front of the integral, and gives the result Arturo mentioned.
 
This correctly gives the positive work that must be done on the positive test charge to move it closer to the positively charged cylinder.  The electric field does negative work, and whatever forces pushes it does positive work.
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02/16/17

Steven W.

tutor
That negative sign on the path integral, by the way comes from the fact that moving along the electric field in the direction it points means going away from positive charges (where electric field lines originate and electric potential is high positive) toward negative charges (where electric field lines end and potential is high negative).  So a path in the direction of the electric field must be moving toward lower potential, and moving against the electric field goes toward higher potential.
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02/16/17

Arturo O.

In addition, the electric field is given by
 
E = -∇V = -(∂V/∂x i + ∂V/∂y j + ∂V/∂z k)
 
Hence, the electric field points naturally in the direction of falling potential; i.e. downhill.  But if you multiply it by a negative charge to get force, the force vector will point uphill.  That is good to keep in mind, that positive charges accelerate downhill and negative charges accelerate uphill, as far as electric potential is concerned.
 
Steven, what was your PhD dissertation in?  I remember you mentioned once you did experimental physics.
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02/17/17

Steven W.

tutor
Yes, Arturo, I was an experimentalist.  I worked on optical instrumentation; specifically, a new kind of Fourier transform spectrometer.
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02/17/17

Arturo O.

Steven,
 
That sounds interesting.  I always liked experimental atomic and molecular spectroscopy.
Report

02/18/17

2 Answers By Expert Tutors

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Arturo O. answered • 02/16/17

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