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Homework Help: Help to understand electrostatics fully

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data

    My Physics course does not give an in-depth overview of electrostatics at all (it is rather shallow, focusing on electronics and circuit analysis). However, I find that I really need to understand it fully to form the basis for magnetism and generation. So... I have a few questions that I wanted to ask:

    1. Electric potential and charge signs - I know that the electric potential (energy) is set to be zero at infinity, i.e. PE = 0 for r = ∞. For a stationary positive charge and a positive test charge, I know that, as distance decreases, potential energy increases...

    But what about:

    • PE of negative charge relative to positive stationary charge? I think it is zero at ∞ but gets negative as distance decreases, as W = -ΔPE so as distance decreases, ΔPE is negative, and W is positive, so speed increases (correct!)
    • PE of negative charge relative to a stationary negative charge should be same scenario as positive/positive?
    • PE of positive charge relative to negative stationary charge should be the same as the neg/pos scenario
    My books (and Khan Academy) only look at the positive/positive case.
    So what about electric potential? I know it depends only on the stationary charge (i.e. the central charge of interest). So what about the signs of potential for a given charge?

    • The potential relative to a positive charge should be positive, and should be negative for a negative charge. This would agree with what I'm figuring from PE (if we put signs into the equation V=kq/r
    2. Signs of potential, and ‘potential across …’ – what does it mean to say ‘there is a potential difference across the cell wall of 1.7 v’ for example? What is the meaning of sign here? Should I take an increase in positive charge concentration as a positive p.d., and a decrease in positive charge concentration as a negative p.d.? Is there such a convention?

    For example:

    “Another interesting biological example dealing with electric potential is found in the cell’s plasma membrane. The membrane sets a cell off from its surroundings and also allows ions to selectively pass in and out of the cell. There is a potential difference across the membrane of about –70 mV”

    What is the meaning of the negative sign of 70 mV? I understand the potential difference of 70 mV as meaning that charge on either side of the membrane will have a difference in potential relative to the other side of magnitude 70mV, but whether it is higher or lower than the other side, how do I know? So if it was magnitude, no problem, I understand what it means. But -70mV? What would that mean? Which side is at higher potential? The cell or the surroundings? Would the negative be meaningless unless extra info on which side was positive/negative was known? (Just a bit confusing)

    3. What do batteries actually do? I know they create a potential difference, hence a potential gradient ⇒ electric field. So this accelerates charges through the circuit and produces energy. Makes sense. But what about capacitors? The battery transfers charge from one plate to another. Now, do the terminals of the battery actually have net charge? I’m pretty sure they don’t, I can’t see any electrostatic effects happening from the negative end of a battery. So why can a battery produce charge separation on a capacitor, when it doesn’t do it at its terminals? Or is the average battery far too weak for this kind of thing?

    4. This question is not on electrostatics, but it is an application of magnetism – the force between two infinite current-carrying wires. (See attached image)

    So what happened here is – the force due to wire 1 on wire 2 is calculated, and apparently that’s it – the force between the two wires is equal to that…. But shouldn’t the force be multiplied by 2 because of the force due to wire 2 on wire 1? Could I have an explanation as to why it is not necessary?

    I don’t get this… L Help please!

    2. Relevant equations

    Voltage : V = kQ/r

    3. The attempt at a solution

    ^^ See above

    I would be very grateful if anyone could help me debunk these confusing points!

    Attached Files:

  2. jcsd
  3. May 13, 2015 #2


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    There is potential (expressed in Volts, which are Joule/Coulomb) and there is potential energy (expressed in Joules).

    Potential and potential energy are determined but for a constant (the only thing you can 'feel' is a derivative). The usual choice is 0 at infinity.

    Here your equation fits in: the electrostatic potential field due to a charge Q at the origin is kQ/r .

    A charge q in a potential V has potential energy qV. If you talk about V you can consider this q a 'test charge'.

    And what you can 'feel' is the electrostatic field ##\vec E = -\vec \nabla V##, or rather the force ##\vec F_e = q\vec E## .
    gradient ##\vec \nabla f = ( {\partial f\over \partial x}, {\partial f \over \partial y}, {\partial f \over \partial z} )##.​

    As distance decreases, electrostatic potential energy increases if and only if charge q and potential V have the same sign -- for the simple case of a charge Q at the origin causing the potential.

    Four yes. All make sense when you consider the factor qQ in the PE.

    2,3,4,5,6, .... sorry, lack of time. Perhaps later if no-one else chips in. (They may be reluctant because PF isn't a physics teaching institution but a homework assistance site :smile:)
  4. May 14, 2015 #3


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  5. May 14, 2015 #4


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    Well, as long as someone is reading on, it probably isn't all wasted.
    On a didactic note: I was triggered also by
    So they didn't find it worthwhile to also do neg/neg, neg/pos and pos/neg. Not from lack of space or time, you may assume -- if you think about it. My impression: Their implicit assumption is that by the time you got this far, you can make that step by yourself. Now, there will be hundreds, if not thousands of readers/viewers who -- while in the learning process -- are not completely overseeing everything just yet and go on through without even realizing there are things to wonder about and things to ask. So: a big compliment to you that you do not take everything for granted, do ask yourself questions and think about what it actually means and what it's all about. I am honestly impressed.

    Good and well, but what to actually do at such a point ? I wish I could answer that ! Sometimes diving in is worthwhile the cost in time and effort, at other times what is unclear and vague half way becomes evident and more concrete further on. In good text books you can have an awful lot of benefit from looking at the exercises early on: once you can see the relationship between the exercises and what's presented in the chapter, you're relatively OK. I'm old now, but what helped me tremendously while a student, was to going to the library and grab a few other (text)books on the same subject: see what I liked most and see what all of them have in common. I suppose nowadays the internet takes the place of the library. But I find there is a lot of distraction and low-quality stuff that wastes a lot of time ...

    (He, is this still PF HW help ? ...)

    So to go on with 2.
    You can take this difference literally and postpone knowing whether it's positive or negative. If it doesn't follow from context and it's not stated, there still are a lot of possibilities:
    • surf around. I didn't find your quote, but page 136 here is very clear:
    The voltage across a membrane, called a membrane potential, ranges from about ##-##50 to ##-##200 millivolts (mV). (The minus sign indicates that the inside of the cell is negative relative to the outside.)
    (Interesting subject, by the way! And going to the net was a lot faster than going to the library :smile:)
    • postpone. Once they say "a positive K+ is pushed out" you know that outside the potential is lower.
    As long as you realize there is some ambiguity, you're all right. And writing absolutely fool-proof and unambiguous is next to impossible -- and long winded -- and boring :wink:, so give the author some slack.

    And 3.
    Having charge is one thing, what batteries do is maintain a voltage, i.e. replenish charge that moves on. By some chemical means. Google.
    Capacitors just give what they have (and discharge).
    And yes, a battery pumps charge to a capacitor until the voltage on both is equal. No voltage difference -- no charge being urged to move any more.

    And 4.
    You got it. From symmetry (turn over an angle ##\pi##), you see that the forces are equal and opposite. The text is actually very clear most of the time: the therm "force between the wires" is being used qualtitatively in the figure caption.
    And I think you did indeed put your finger on a weak spot in the last line ! Instead of the vague term "between" they should have said "on each of the wires carrying the" .
    Now go looking for an accompanying exercise at the end of the chapter and see if you nevertheless get the same answer they get :wink:


  6. May 14, 2015 #5


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    I read it through (quickly) and might do so again later :sorry: (that's why being an educator is so fustrating
    - is anybody actually listening - If you care they will.)

    Note: The sign of a potential tells you in which direction the (+) current will be.
    Current goes from a higher to a lower potential.
    Last edited: May 14, 2015
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