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Definition of electric potential in case of negative charges

  1. Jun 30, 2010 #1
    Hello, I just read on "Integrated electronics" by Millman the following definition:

    "The potential V (volts) of points B with respect to point A is the work done against the field in taking a unit positive charge from A to B.

    For a one-dimensional problem with A at [itex]x_0[/itex] and B at an arbitrary distance x, it follows that:

    [tex]V = - \int_{x_0}^{x} E dx [/tex]

    where E represents the X component of the field.

    Differentiating the above equation we have: [tex]E = - \frac{dV}{dx}[/tex]

    The minus sign shows that the electric field is directed from the region of higher potential to the region of lower potential."

    I'm wondering if the above concepts (like the one regarding the direction of the electric field) hold in the following cases:
    - the charge taken from A to B is an electron;
    - the charge is an electron, plus A is grounded and B has a negative voltage;


    Any help would be appreciated. Thanks in advance.
     
  2. jcsd
  3. Jun 30, 2010 #2

    jtbell

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    Staff: Mentor

    The potential (V) depends only on position, and is the same regardless of whether the charge at that position is positive or negative, or indeed whether there is an actual charge there at all.

    The potential energy (PE) of a charge located at a certain position depends on whether the charge is positive or negative, because PE = qV and you have to take into account the signs of both q and V.
     
  4. Jun 30, 2010 #3
    Let's start with Coulomb's law:

    [tex]\overrightarrow{F} = k_\mathrm{e} \frac{q_1q_2}{r^2} \overrightarrow{r}[/tex]

    We know that if the two charges have same sign they reppel each other because the force becomes positive. This happens because the vector r is a vector that has the direction of the line that connects the two charges, of course, but it also goes outward of the charge that you are not measuring the force on. If we want to know the force on Q then the vector goes in the direction that goes outwards of q.

    Q+ q+
    . - - - - .---->F and r

    Look, F (force acting on q) and r (vector that connects Q and q and goes outward of Q).

    Let's assume a positive test charge q. Then we can deduce the electric field:

    [tex]\mathbf{E}=\lim_{q \to 0}\frac{\mathbf{F}}{q}[/tex]

    We take the limit because if we insert any charge in the system of charges it will make they move and the force on test charge will change too.

    [tex]\mathbf{E}= k_{e}{Q \over r^2}\mathbf{\hat{r}}[/tex]

    Now we know that if our initial charge Q is positive the electric field due to this charge in every point goes in the direction that outwards the charge. Thus if we have a positive charge the electric field due to it is always going outwards of the charge (and if the charge is negative the electric field goes in it). So here is the answer to your first question:

    The electric field of negative charges goes inwards the charge.

    Now the electric potential is the work done when moving a charge from A to B.

    [tex]V = - \int_{P1}^{P2} E dx[/tex]

    So about the minus sign think: what should spend more energy (assuming we have two charges q+ and Q+): bringing q towards Q or taking q backwards Q? Of course the answer is: bringing q towards Q, because if we do that we are trying to go in the opposite direction of Q's electric field (so in this case the work done must be positive). That's the reason of the minus sign in the potential's formula.

    So if the charge that we are bringing to Q is an electron we take negative work. We take positive work only if we try to pull the electron away from the Q charge (which is positive). That's the answer to your second question.

    Now suppose we have two points: A grounded (0 potential) and B (negative potential). We already know that positive charges tends to go from higher potentials to low potentials (and negative charges from low potentials to higher potentials). The electron, thus, will go from B to A because the potential in B is lower than the potential in A.

    Am I right?

    I hope it helps you...

    Rafael Andreatta
     
  5. Jul 4, 2010 #4

    Thank you, I really appreciated your explanation! There's just one thing I can't get yet: the Millman's book introduces the concept of a potential-energy barrier by an example where A (grounded) and B (negative potential) are two parallel plates and an electron moves from A to B. I got the main concept.. but can an electron move from A to B? How is that possible?

    Here's the configuration:
    [PLAIN]http://img341.imageshack.us/img341/7316/picture9mr.png [Broken]

    Thank you.
     
    Last edited by a moderator: May 4, 2017
  6. Jul 4, 2010 #5
    Sorry, I have never seem this concept before... I will let some other person help you this time...
     
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