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Electric field from map of voltage potentials

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data

    I recently retired and thought it would be fun to reread my old EM texts. I took Fields and Waves 35 years ago for my undergraduate electrical engineering degree, so I am more than a little rusty. This question is for my personal knowledge, not for a class, but seems to best fit in the homework category.

    I am trying to figure out the following example from Hayt, Engineering Electromagnetics (3d ed. 1974 McGraw-Hill) I have added the A, B, and C labels.

    Hayt1.jpg
    __________________

    The example is in Chapter 4, Energy and Potential. For context, the chapter discusses the energy expended in moving a point charge in an electric field (section 4.1); the line integral (section 4.2) (including the work from moving a charge in a radial path in the field of a infinite line charge); the definition of potential difference and potential (section 4.3); the potential field of a point charge (section 4.4); the potential energy of a system of charges: conservative property (section 4.5); potential gradient (section 4.6); the dipole (section 4.7); and energy density in the electrostatic field (section 4.8). This example occurs in the section on potential gradient (section 4.6), which defines the del vector operator [∇] and the relationship between E and V ([itex]{\vec E= -∇{\vec V}}[/itex]). Poisson’s and Laplaces’s equations are not introduced until Chapter 7.

    Because the problem states that this is a portion of a two-dimensional electric field and the potential surfaces are concentric, I initially assumed that we are dealing with an electric field from an infinite line charge and that the electric field is only radial. As explained in the attempt at a solution, I now think this may be wrong because the voltage increases with increasing radius.

    Because the potential decreases with decreasing radius, the electric field must be radially inward, and I assumed that the line charge is negative.

    2. Relevant equations
    For an infinite line charge pL (C/m) and a radial field:

    [itex]E = \frac {ρ_L}{2πε_0r}[/itex]

    [itex]{V_{AB} = \int_B ^A \vec E\, \cdot \, d\vec L = \int_B ^A \vec E\, \cdot \, d\vec r} [/itex]

    The equation for a circle:

    [itex]x^2+y^2=r^2[/itex]

    3. The attempt at a solution

    I used the equation for the voltage difference going from point B (on the 104 V circle) to point A (point a on the 106 V circle) in the radial direction, which I choose because the points are shown as 2 mm = 0.002 m apart:

    [itex]{V_{AB} = V_A - V_B = - \int_B ^A \vec E_r\, \cdot \, d\vec r = - \frac {ρ_L}{2πε_0} ln(r) |_{r_B} ^{r_A}} [/itex]

    [itex]{\ \ \ \ \ \ \ \ 106-104 = - \frac {ρ_L}{2πε_0} ln(\frac {r_A} {r_B})} [/itex]

    [itex]{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 = - \frac {ρ_L}{2πε_0} ln(\frac {r_A} {r_B})} \ \ \ \ \ \ [/itex] Eqn. (1)

    And, from the map

    [itex]r_B = r_A - 0.002[/itex]

    There are 2 variables, [itex]{ρ_L}[/itex] and [itex]{r_A}[/itex], and we have only one equation. Picking another voltage difference and going from point C (on the 102 V circle) to point A (point a on the 106 V circle) in the radial direction (it doesn't seem that this is right because the distance has to be estimated, unlike the distance from B to A).

    [itex]{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 = - \frac {ρ_L}{2πε_0} ln(\frac {r_A} {r_C})}\ \ \ \ \ \ [/itex] Eqn. (2)

    And, from the map, point C is about 3.7 mm from point A

    [itex]r_C = r_B - 0.0037[/itex]

    Substituting and dividing Eqn. (1) by Eqn. (2):

    [itex]{\frac{2}{4} = \frac {1}{2} = \frac {ln(\frac {r_A}{r_A - 0.002})} {ln(\frac {r_A}{r_A - 0.0037})} } [/itex]

    [itex]{ln({\frac {r_A}{r_A - 0.002})} = 2{ln(\frac {r_A}{r_A - 0.0037})} } [/itex]

    [itex]{\frac {r_A}{r_A - 0.002} = e^2({\frac {r_A}{r_A - 0.0037})} = 7.3891({\frac {r_A}{r_A - 0.0037})}} [/itex]

    [itex]{r_A - 0.002 = 7.3891(r_A - 0.0037)} [/itex]

    [itex]{r_A = 0.004} [/itex]

    Unfortunately, this does not agree with the map. The first problem I see is that the distance between every 2 V equipotential surface is decreasing towards the center (e.g., from 106 V to 104 V is 2 mm and from 104 V to 102 V is about 1.75 mm), so the radius from the center to 106 V should be no more than 106/2 = 53 mm = 0.053 m and the radius to 104 V should be 2 mm less than that = 0.051 mm (just to get a subjective feel for what is happening). This would mean that ln(r)<0, ln(0.051)= -2.9754, ln(0.053)= -2.9375, and therefore ln(0.051)>ln(0.053) which implies that the voltage potential would decrease with increasing radius, contrary to the figure. If r>1, then ln(r)>1 and is increasing, e.g., ln(1.02)=0.0198, ln(1.04)=0.0392, and ln(1.06)=0.0583; however, r is not greater than 1.

    The second problem I see is that it would seem that for a negative infinite line charge, the values of equipotential should be decreasing with increasing radius, e.g., -104 V, -102V, -100 V, etc., so V##\to##0 as r##\to \infty##. However, the figure shows positive increasing values. Similarly, for a positive line charge, the values of the equipotential would be expected to decrease with increasing radius, e.g., 104 V, 102 V, 100V, etc., so V##\to##0 as r##\to \infty##. However, the figure shows increasing values.

    It seems, therefore, that the usual relationships for an infinite line charge are not to be used and the problem may be asking to find a function V(x,y). Because the equipotentials are concentric, it seems that the equation for a circle is at least part of the answer. For point a it seems that we only need to worry about a function in in terms of the radius, V(r). It seems that there must be a factor of ln(r) somewhere to account for the fact that the distance between equipotential surfaces is increasing with r. And it seems that r>1 because the voltage potential is increasing with r. I don’t see an easy way to derive a function that gives the values of V or the value of the radius.

    I have tried working backward from the solutions, but this doesn't seem to work. From part (a):

    [itex]{\vec E = -900 \vec a_y = -900 \vec a_r = -∇{\vec V} = \frac {\partial V}{\partial r} \vec a_r}\,\,\,\,\,[/itex] (where [itex]\vec a_y = \vec a_r[/itex])

    [itex]{V = \int \vec E\, \cdot \, dr\vec a_r = \int 900 \vec a_r \cdot dr\vec a_r = 900r}[/itex]

    This doesn't help in terms of giving a general expression for V(x,y) or V(r) that I can see and, of course, I needed E to get this far. From part (b), the gradient is along the radius, with x=.006 from center, y=unknown. so I tried making the gradient components proportional to the distances: .006/450=y/700. This gave a value of y=.006(700/450)=0.0093. This does not seem to agree with the map. However, the value for [itex]{\vec E} [/itex] at point b is [itex]{-\sqrt {450^2 + 700^2} = -832 \vec r} [/itex] which is consistent with [itex]{\vec E = -900 \vec r} [/itex] for point a (since E is increasing with decreasing radius).

    Because the example occurs early in the book, and the answers look simple, it does not seem that it should be too hard. Probably I am overlooking something obvious. Any hints would be appreciated.
     
  2. jcsd
  3. Nov 3, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    Hello LeeB66. Welcome to PF!

    As you noted, it does not look like this is the potential distribution from a line charge.

    Perhaps they want you to get an approximate answer by approximating ##\vec{E} = -\vec{\nabla}V##.

    Thus, ##E_x = -\frac{\partial V}{\partial x} \approx -\frac{\Delta V}{\Delta x}## while keeping ##y## constant. Similarly for ##E_y##.

    You will only be able to get approximate results, say to the nearest 50 V/m (if you're lucky :smile:).
     
  4. Nov 3, 2015 #3

    Merlin3189

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    Gold Member

    I have no idea what you are supposed to do and I haven't yet worked out my answer, but when I read the question, before I looked at your attempts, I assumed it was to be done by plotting graphs. I didn't read anything about symmetry and assumed it was an arbitrary field.

    Linear interpolation gave me very approximately
    ya = - 4V / 4.4mm = -0.091 V/mm = - 910 V/m
    xa = 0
    yb = -2V / 2.8mm = -0.71 V/mm = - 710 V/m
    xb = -2V / 4.3mm = -0.47 V/mm = - 470 V/m

    Perhaps if I plot the graphs and draw tangents I might get a more accurate result, but since I read the positions from the diagram, it wont be much better.`

    Edit Having read the answer, I had just put -ve signs in for the PD's and simply took the contours as 2V differences without looking at the actual values. Looking again at the diagram to plot the graphs I can't see the reason for getting a negative answer, so I would not now get -ve answers and although the magnitude of my answers might be approximately right, their sign would be wrong.
     
    Last edited: Nov 3, 2015
  5. Nov 3, 2015 #4
    Thanks to both of you. The answers looks so exact that I focused on finding an exact mathematical solution rather than trying to interpolate from the map. The fact that the line appear concentric and the 102 V and 104 V curves appear to be exactly 2 mm apart seemed like a clue. However, I think Merlin3189 is correct that the solution is to interpolate. This is probably why there is such a detailed map of the potentials. This simplifies the solution, which is in keeping with the problem being early in the book, and avoids any need to assume symmetry. I will consider the problem solved. Again, thanks to both of you for spending the time.
     
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