Electrostatic Boundary Conditions

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kmm
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In Griffith's section about electrostatic boundary conditions, he says that given a surface with charge density [itex] \sigma [/itex], and take a wafer-thin Gaussian pillbox extending over the top and bottom of the surface, Gauss's law states that: [tex] \oint_{S} \mathbf{E} \cdot d \mathbf{a} = \frac{1}{\epsilon_{0}} Q_{enc} = \frac{1}{\epsilon_{0}} \sigma A [/tex] Now, in the limit that the thickness of the pillbox goes to zero, we have: [tex] E_{above}^{\perp} - E_{below}^{\perp} = \frac{1}{\epsilon_{0}} \sigma [/tex] The image he gives is attached in this post. He says for consistency to let upward be the positive direction for both, but I don't understand why he has E pointing up above and also up below the surface. I would think E is pointing up above the surface and down below the surface so that when we take [tex] \oint_{S} \mathbf{E} \cdot d \mathbf{a} [/tex] we would actually get [tex] E_{above}^{\perp} + E_{below}^{\perp} = \frac{1}{\epsilon_{0}} \sigma [/tex] getting a plus instead of minus since [itex] \mathbf{E}_{below} [/itex] and [itex] d \mathbf{a} [/itex] both point down canceling the negatives. What am I thinking wrong?
 

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  • #2
Meir Achuz
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'let upward be positive for both directions,' is just a sign convention. The values of E_above and E_below
can be positive or negative numbers, as can sigma.
 
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  • #3
kmm
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'let upward be positive for both directions,' is just a sign convention. The values of E_above and E_below
can be positive or negative numbers, as can sigma.
OK, but when I evaluate the vectors in the integral, [itex] \mathbf{E} \cdot d \mathbf{a} [/itex] for the top surface comes out positive since [itex] \mathbf{E} [/itex] and [itex] d \mathbf{a} [/itex] both point up. On the bottom surface [itex] \mathbf{E} [/itex] and [itex] d \mathbf{a} [/itex] both point down giving a positive value also. Unless below the surface [itex] d \mathbf{a} [/itex] actually points up.
 
  • #4
kmm
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'let upward be positive for both directions,' is just a sign convention. The values of E_above and E_below
can be positive or negative numbers, as can sigma.
Actually, I think I understand now. You're saying we don't know whether the charge is positive or negative, so we don't actually know what direction E points in for the top or bottom. The negative sign in front of E_below would then come from [itex] d \mathbf{a} [/itex] pointing down. Depending on the charge, we could then put a positive or negative value in for E_below or E_above. Is that correct?
 
  • #5
Meir Achuz
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Not quite. If there is also an E field in addition to that from the surface charge, E_above and E_below could each have either sign, but their difference wohld be given by sigma.
 
  • #6
kmm
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Not quite. If there is also an E field in addition to that from the surface charge, E_above and E_below could each have either sign, but their difference wohld be given by sigma.
That doesn't seem right because any additional field other than that due to the surface charge would have to be due to charge outside the pillbox and would therefore not contribute to the flux. This calculation doesn't consider anything that doesn't contribute to the flux which is why the pill box thickness is shrunk to zero. If [itex] \sigma [/itex] is positive or negative, in either case E field on either side of the surface will be pointing in different directions, and Gauss's law always gives [itex] \frac{\sigma}{\epsilon_{0}} [/itex].
 
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  • #7
jtbell
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Consider the field as having two parts: the field due to the surface charge, ##\vec E_{surf}##, and a field due to external sources ##\vec E_{ext}##. For simplicity let the surface be horizontal and positively charged, and let ##\vec E_{ext}## be vertical upwards.

First suppose ##\vec E_{ext} = 0##. We know from Gauss’s Law applied to ##\vec E_{surf}## alone, that it’s upwards above the surface and downwards below the surface, and has magnitude
$$E_{surf} = \frac{\sigma}{2 \epsilon_0}$$

Now suppose ##\vec E_{ext} \ne 0##. The net field, ##\vec E = \vec E_{ext} + \vec E_{surf}##, has different magnitudes above and below the surface. Take the case that ##E_{ext} > E_{surf}##. Then ##\vec E## is upwards on both sides of the surface and
$$E_{above} = E_{ext} + E_{surf}\\
E_{below} = E_{ext} - E_{surf}$$

Find the difference between these two magnitudes, and connect it to our value of ##E_{surf}##.

Repeat for the case ##E_{ext} < E_{surf}##, if you like.
 
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  • #8
kmm
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Consider the field as having two parts: the field due to the surface charge, ##\vec E_{surf}##, and a field due to external sources ##\vec E_{ext}##. For simplicity let the surface be horizontal and positively charged, and let ##\vec E_{ext}## be vertical upwards.

First suppose ##\vec E_{ext} = 0##. We know from Gauss’s Law applied to ##\vec E_{surf}## alone, that it’s upwards above the surface and downwards below the surface, and has magnitude
$$E_{surf} = \frac{\sigma}{2 \epsilon_0}$$

Now suppose ##\vec E_{ext} \ne 0##. The net field, ##\vec E = \vec E_{ext} + \vec E_{surf}##, has different magnitudes above and below the surface. Take the case that ##E_{ext} > E_{surf}##. Then ##\vec E## is upwards on both sides of the surface and
$$E_{above} = E_{ext} + E_{surf}\\
E_{below} = E_{ext} - E_{surf}$$

Find the difference between these two magnitudes, and connect it to our value of ##E_{surf}##.

Repeat for the case ##E_{ext} < E_{surf}##, if you like.
In either case, when I take [itex] E_{above} - E_{below} [/itex] I get [itex] 2E_{surf} = \frac{\sigma}{\epsilon_{0}} [/itex] so the external fields cancel. I think my confusion in the OP has been solved. I didn't realize that Griffith's was being general with [itex] E_{below}^{\perp} [/itex] and [itex] E_{above}^{\perp} [/itex] (ie either one could be positive or negative).
 

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