Electric field between two cylinders in a vial

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CMLeo
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Homework Statement



Hello everyone,

I am writing here regarding a doubt I have about electric fields. Our set up consists on two rod shape electrodes in a cylindrical glass vessel separated by 6 mm from each other, with one of the electrodes grounded (electrodes dimensions: 0.5 mm diameter and 10 cm long). As far as I know the electric field between the electrodes is constant for uniform DC fields, but we have here an inhomogeneous field and I am not quite sure how to calculate the electric field gradient. A voltage up to 400 V is supplied within the system (dielectric solvent, so no current is induced and no chemical reaction is expected).

Two side-by-side diagrams: a top view of two rod electrodes in a glass vial with field lines between them, and a front view of two parallel plates separated by distance d with dashed arrows showing a uniform field, regions labelled 1, 2 and 3.

Homework Equations



Calculate the electric field and voltage in 1, 2, and 3 as a function of the distance.

The Attempt at a Solution


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The voltage and electric field of a charge cylinder as a function of the distance is given by:

E(r)= λ/(2πεr) r≥a
V(r)= (λ ln(a/r))/(2πε) r≥a

where λ is the charge per unit length, ε is the permittivity of the free space, a is the radius of the cylinder, and R is the distance from one of the electrodes. V=400 V at the charged electrode and 0 V at the grounded electrode.

First, I think I should replace the permittivity of the free space for the medium (ε'=ε x εm). For a point co-linear with the centres of each cylinder we have an electric field in 2 equal to:
Three equations for the electric field of two line charges: E_left(r) = λ/(2πε′r) r̂, E_right(r) = −λ/(2πε′(d−r)) (−r̂), and their sum E(r) = λ/(2πε′)(1/r + 1/(d−r)) r̂.


At 1:
Equation E(r) = (λ / (2πε′)) × (1/(−r) + 1/((d + r))) times unit vector r-hat, giving the electric field at a point between two line charges separated by distance d.


At 3:
Equation E(r) = (λ / (2πε r)) (1/r − 1/(d + r)) r-hat, with λ, ε, r and d in the expression.


For the potential difference between the electrodes:

At 2:
Equation: delta V = integral of E dl = (lambda / (2 pi epsilon)) times integral of (1/r + 1/(d-r)) dr, which equals (lambda / (2 pi epsilon)) times [ln(r) - ln(d-r)].


At 1 and 3, respectively:
Two equations for the potential difference between cylinders: delta V = (lambda / (2 pi epsilon)) [ln(d + r) - ln(r)] and delta V = (lambda / (2 pi epsilon)) [ln(r) - ln(d + r)].


I am not sure whether this attempt is right or not. Besides, I would like to have the electric field without the charge per unit length since I do not know how to calculate with the information I have. So, I thought in obtain λ as a function of the voltage (I have the supply voltage) and replace it in the E equation. However, I do not know if this would work since the voltage at the second electrode is 0 V, should I assume that beyond the second electrode I have 0 V and the walls of the vial are 0 V as well?

Thanks in advance for any help.
 
Attachments
  • Two diagrams: left, top view of a charged electrode and a grounded electrode, each radius a, inside a glass vial with dielectric solvent, with dashed field lines and arrows; right, front view showing regions 1, 2 and 3 with horizontal dashed arrows between a charged plate on the left and a grey plate on the right.
    upload_2018-1-15_9-29-15.png
    12 KB · Views: 773
  • Two side-by-side diagrams: a top view of two rod electrodes in a glass vial with field lines between them, and a front view of two parallel plates separated by distance d with dashed arrows showing a uniform field, regions labelled 1, 2 and 3.
    upload_2018-1-15_9-53-25.png
    11.7 KB · Views: 2,130
  • Equation E(r) = λ/(2πε') times (1/r + 1/(d − r)), giving the electric field at distance r between two cylinders separated by d.
    upload_2018-1-15_9-55-33.png
    892 bytes · Views: 675
  • Equation E(r) = λ/(2πε') times (1/r + 1/(d + r)), giving the electric field at a point between two line charges separated by distance d.
    upload_2018-1-15_9-59-14.png
    890 bytes · Views: 709
  • Three equations for the electric field of two line charges: E_left(r) = λ/(2πε′r) r̂, E_right(r) = −λ/(2πε′(d−r)) (−r̂), and their sum E(r) = λ/(2πε′)(1/r + 1/(d−r)) r̂.
    upload_2018-1-15_10-16-40.png
    1.7 KB · Views: 1,194
  • Equation E(r) = (λ / (2πε′)) × (1/(−r) + 1/((d + r))) times unit vector r-hat, giving the electric field at a point between two line charges separated by distance d.
    upload_2018-1-15_10-17-41.png
    981 bytes · Views: 1,134
  • Equation E(r) = (λ / (2πε r)) (1/r − 1/(d + r)) r-hat, with λ, ε, r and d in the expression.
    upload_2018-1-15_10-19-30.png
    984 bytes · Views: 1,076
  • Equation: delta V = integral of E dl = (lambda / (2 pi epsilon)) times integral of (1/r + 1/(d-r)) dr, which equals (lambda / (2 pi epsilon)) times [ln(r) - ln(d-r)].
    upload_2018-1-15_10-48-24.png
    2 KB · Views: 1,133
  • Two equations for the potential difference between cylinders: delta V = (lambda / (2 pi epsilon)) [ln(d + r) - ln(r)] and delta V = (lambda / (2 pi epsilon)) [ln(r) - ln(d + r)].
    upload_2018-1-15_10-53-5.png
    1.1 KB · Views: 1,120
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Welcome to PF.

The distribution or movement of charges outside the vial will connect to the internal electrodes through the glass wall, it is a glass dielectric capacitor. Should you now consider the outside surface of the vial to be an equipotential ground?

You need to identify the boundary condition. Is either vial surface an equipotential, or is the glass a series capacitance to the unspecified external environment. There are three dielectrics involved, the liquid, the glass and the air. The ratio of dielectric constant of the liquid to the glass will determine relative field gradient.
 
I would say that the glass is a equipotential ground. The air should not be consider, I assumed that the electrode surface is complete immerse within the dielectric solvent.
Regarding the boundary conditions, I am not what else I can consider to establish them. Any advise?

Thanks.
 
The problem is that glass is a dielectric insulator. One solution I can see would be to wind a helix of thin wire around the outside of the vial and connect it to ground. That would define the external glass surface as the ground potential boundary condition. In the real world I would also paint the external glass surface with ethylene glycol = radiator anti-freeze, which is hygroscopic so it will remain conductive and reduce the sensitivity to external electric fields.

The capacitance or electric field model will will need to include the glass and the liquid, but the asymmetric geometry makes it difficult. You might simplify the model by considering the two electrodes at symmetrical voltages ±V/2, about the external ground. Also, if the glass and liquid dielectric constants were of similar value, then the inner glass surface would "disappear" and so simplify the model.

You might look at the capacitance modelling of transmission lines such as coaxial cable, or more specifically the model for "twinax" screened cable.

I am not sure if you are presenting a set homework problem, or designing a small part of some experimental equipment.