Electric Potential & UE Lesson 02

Hey everyone... power's out here in Kitsap. I have my generator fired up and I'm camped out around my space heater. I do have limited internet access through my phone, but I can't do Hangouts today.

Please review the following lesson plan... the partial derivative stuff may be kinda fun, but that's more of an aside than anything we have to actually get to (if you look in the back of your calc book you'll see calculus with multiple derivatives and integrals <hence the term multi-variant calculus>. That is usually taught as 3rd semester calc.)

Finally -- I can't emphasize how important it is to keep up with the work. We are NOT going to spend ANY time catching up whenever it is that we return to class. Also, there's a very good chance we'll have a test THIS week if we ever make it back... so stay frosty (so to speak)

OPENING QUESTIONS: Electrical potential and electrical potential energy ain't not the same thing.... why not?


1) I will be able to calculate electrical potential by analyzing the contribution of a few charged particles after today's class.

2) I will be able to calculate electrical potential by analyzing a continuous series of charged particles (using integration) after today's class.

3) I'm going to get my first real look at partial derivatives and how to apply them in tonight's homework.




  • Test Charge: A mathematical construct-- a charge that does not exert any influence on surrounding particles but IS influenced by other electric fields.
  • conductor - materials where electrons can roam
  • insulator - materials that keep their electrons close to home
  • coulomb - a unit of electrical charge (see below)


  • e-mass = 9.1 x 10-31 kg
  • charge of an electron = 1.60 x 10-19 coulombs (C)
  • ke = 8.987 x 109 Nm2/C2
  • ke = 1/4πεo


      • Electric Potential (V) and Electric Potential Energy (UE)
        • U = -qE ds
        • Vb - Va = ∆V = U/q = -E ds
        • W = qV = -UE
        • V = keqi/ri (between some point in space and each charged particle present in that region of space)
        • UE = keq1q2/ri (between each particle in space and every other particle in that region of space)
      • Electrical Force:
        • Fe = (keq1q2)/r2
        • Fe=qE
      • Electrical Field:
        • E = Fe / qo where qo  = a positive test charge
        • E= (keq)/r2
      • Gauss's Law:

        Φ = E dA = Ecos dA = qin/εo

        Φ = qin/εo : the electric flux through an ENTIRE gaussian surface is equal to the algebraic sum of the charges INSIDE the surface divided by the permitivity of free space


    Spend a few moments familiarizing yourself with the relationships between U, V and such...this needs to be right on the tip of your tongue... NOW!


    There are two important considerations in section 25.3:

    The total electric potential between multiple charged particles is equal to the sum of the electric potential of those charged particles:

    V = ke∑qi/r

    Similarly, the total electrical potential energy (UE) between charged particles is the sum of the potential energy between each pair of charged particles:

    U = ke∑q1q1/r

    Note electric potential is a scalar quantity. As such we are not interested in vectorizing "r".

    Remember: it takes WORK (adding energy) to make a charged particle move 'up' stream. Consequently when a charged particle moves "downstream" in an electric field, the system as a whole LOSES energy as the particle goes from a higher to lower potential.

    Please take a look at example 25.3

    Let's discuss 25.3 part B.... it's a tad convoluted.


    (FUN WITH NUMBERS!!!! (hold on to your hats folks! We *may* get to this at the VERY end of the term, but for now I'd like you to take a first look at it-- please do NOT stress over this)

    Recollect (?) that:

    Vb - Va = ∆V = U/q = -E ds

    since we often replace with 'd' when things are getting interesting -- let's simplify that a bit:

    dV = -E ds

    let's simplify that further for the case of a uniform electric field:

    dV = -E∫ ds

    or even more simply:

    dV/ds = -E

    Since s is a generic distance term, let's rewrite that using our generic 'r' for displacement:

    dV/dr = -E

    however, if we consider that the 'r' vector is often rewritten in terms of the x, y and z vectors we can rewrite that as the vector sum of the partial derivatives:

    -E = V/x + V/dy + V/dz

    so...let's say we can describe the electric potential due to an object by the equation:

    E(r) = 2x + 3xy + 4z

    E(x,y,z) = 2x + 3xy + 4z

    so it follows that:

    V/x = 2+3y

    V/dy = 3x

    V/dz = 4


    Now... back to the task at hand--

    If we are considering only the x direction we have just a piece of that:

    -E = V/x

    BUT WAIT... recall that Fe = qE, so what happens if we multiply both sides by q:

    -Eq = q∂V/x

    HOLD THE BUS!!!! qV = U and Eq = Fe so let's substitute!

    -Fe = U/x

    which is an awful lot like: our old friend:

    -Fx = U/x


    We can't *always* deal with the electric potential from a countable number of charged particles. In otherwords, we sometimes need to deal with a continuous number of charged particles.... and that means calculus!

    We often begin with the VERY small:

    We can evaluate a tiny, tiny, tiny bit of electric potential:

    dV = (Ke)dq/r

    And then use integration to sum up that tiny bit of electric potential over MANY charges:

    dV = (Ke)dq/r

    which goes to:

    V = Kedq/r

    In order to make THAT particular fish fry, we will need to get q in terms of r... let's take a look at our favorite long, thin wire (example 25.7)


    1) Review example 25.3 parts a and b. Please make sure that you can MAKE SENSE out of part b. The brute force method works there too (find initial potential E and then subtract that from the FINAL potential E)

    2) WORK example 25.7 especially the 'crossover' using λ. Please don't sweat the integral, remember you can look that up.

    3) Homework problem 25.12

    4) Homework problem 25.44... PLEASE start using integration and DO NOT just brush through it like the author does.