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The question state that 2 large parallel plates are a distance $d$ apart and the field at $d/2$ is $E$ if the distance between the plates are reduced to $\frac{d}{2}$ what is the field strength halfway between the 2 plates?

I have found 2 possible ways 2 solve this online

$E=\frac{V}{d}$ and another formula $E=\frac{\sigma}{2\epsilon_0}$ in the last equation the distance between the plates does not factor.

Why are there 2 ways? this seems to be contradictory and I am not sure what to do.

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  • $\begingroup$ Is there any reason these are contradictory? What do you know about the definition of V? And can't E have two simultaneous relationships? $\endgroup$ – Bill N Mar 21 '16 at 14:53
  • $\begingroup$ V is the potential difference. What I do not get is that one equation is not dependant on the distance between the plates while the other is please help me to understand that $\endgroup$ – Gobabis Mar 21 '16 at 14:56
  • $\begingroup$ Related: physics.stackexchange.com/q/65191/2451 and links therein. $\endgroup$ – Qmechanic Mar 21 '16 at 15:15
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You are confusing the relationship of $V$ to $E$ with the definition of $V$.

The E-field at a point in space depends on the distribution of charge around that point. That's the primary determining factor. The potential difference between the plates (or between two points in space) is defined based on what the E-field is : $$V_{ab}=\int\limits_{r_a}^{r_b}\vec{E}(\vec{r})\cdot d\vec{r}$$

Now, you have to apply this to your specific geometry (small gap between two parallel plates). Remember that the E-field depends on where the charges are. How does the $E$-field vary close to a sheet of uniformly-distributed charge?

The $V$ (potential difference) depends on how the $E$ varies between the two points and how far apart the charges are.

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  • $\begingroup$ I apologise but can you explain this -You are confusing the relationship of V to E with the definition of V- more, for some reason I just struggle with electricity related physics $\endgroup$ – Gobabis Mar 21 '16 at 15:12
  • $\begingroup$ The formula $E=V/d$ is not a definition of $E$. It is a relationship which comes from the definition of $V$ when $E$ is a constant. If $E$ is not constant, that formula doesn't work. $\endgroup$ – Bill N Mar 21 '16 at 15:25
  • $\begingroup$ exactly! when do i use what formula? is there any guideline? $\endgroup$ – Gobabis Mar 21 '16 at 15:28
  • $\begingroup$ See my question at the end of the next-to-last paragraph. That's the key to your question. The guideline is asking yourself "how does the $E$-field change?" OTOH, if you know a function for $V(\vec{R})$ which doesn't have $E$ in it, you can take a spacial derivative (called the gradient) and find the function for $E$. Remember that charge distributio determines $E$, and $E$ integrated over distance determine $V$, unless something like a battery is forcing $V$ to remain constant. In that case, the battery is changing the charge distribution. $\endgroup$ – Bill N Mar 21 '16 at 15:36
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The exact formula to calculate the electric field at a distance $z$ from the centre of a disk of radius $R$ is given at

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

enter image description here

As you can see for $R\gg z$ the magnitude of electric field is constant and given by $E=\frac{\sigma }{2\varepsilon_{0}}$.

The $V=Ed$ formula can be applied to the case where two parallel plates kept at voltage $V$ (external) and separated by distance $d$. See the animation below from

http://www.regentsprep.org/Regents/physics/phys03/aparplate/

enter image description here

As you can see if the potential is constant as the distance gets smaller the electric field increases. If you want to apply the $E=\frac{\sigma }{2\varepsilon_{0}}$ formula here you need to calculate a new $\sigma$ for each $d$ because in this case $\sigma$ is not constant, it increases as the plates come closer as illustrated in the animation by more $+$ and $-$ charges on the plates.

Edit: Answers to the questions in the comments.

Question: What is $\sigma$ and why it increases as the two plates come together under a constant external potential $V$?

Answer: $\sigma$ is a measure of charge density. It can be calculated as total charge divided by total area. $\sigma$ increases as the plates come closer because the charges on each plate can attract more of the opposite charge to the other plate.

Question: How would one calculate the new electric field if the distance between the plates is reduced but there is no external voltage, that is the plates has constant $\sigma$?

Answer: There are two ways. If the $R\gg z$ case is valid then irrespective of the distance between the plates the electric field is constant and given by $E=\frac{\sigma }{\varepsilon_{0}}$.

If $R\gg z$ is not valid then one needs to use

$$E_{z}=\frac{\sigma}{2\varepsilon_{0}}\left ( 1-\frac{z}{\sqrt{z^{2}+R^{2}}} \right )$$

to calculate the electric field.

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  • $\begingroup$ I know i am missing a small piece of the puzzle, or having a stupid mistake in my reasoning, but I just cannot figure out what it is $\sigma$ is $charge/m^2$ how can this increase when the distance shorten? $\endgroup$ – Gobabis Mar 21 '16 at 15:22
  • $\begingroup$ What happens if there isn't a battery between the two plates to hold the potential difference constant? The charge can't change in that situation. $\endgroup$ – Bill N Mar 21 '16 at 15:29
  • $\begingroup$ I have edited the answer to incorporate your questions. $\endgroup$ – physicopath Mar 22 '16 at 8:41

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