Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Uniform Electric Field and Finding Electric Potential

  1. Sep 6, 2012 #1
    1. The problem statement, all variables and given/known data

    7) A uniform electric field of magnitude of E=5.5 ×10^3 N/C points in the positive x axis
    direction. The electric potential at the origin (x,y)=(0,0) of the xy-coordinate system is 0V. Find the electric potential at the point (x,y)=(10.0 cm, 10.0 cm).

    A) -777.8 V
    B) 777.8 V
    C) 550 V
    D) -550 V
    E) 1100 V

    2. Relevant equations

    E = kq/r^2
    V = kq/r

    3. The attempt at a solution

    I thought I would have to find the charge (even though it does not mention any point charges) at (10,10).

    5.5*10^3 N/C = (8.99*10^9 Nm^2/C^2)(q1)

    q1 = 6.1179*10^-9 C

    Then I thought "Oh...well I'm assuming this is the charge at (10,10)."

    V = (8.99*10^9 Nm^2/C^2)(6.1179*10^-9C) / (200^.5) = 3.813 V
  2. jcsd
  3. Sep 6, 2012 #2


    User Avatar

    Staff: Mentor

    Imagine two vertical plates of a parallel-plate capacitor positioned along the x-axis. If the plates are 1m apart, then the potential between them is 5.5x10^3 volts (to fit the specs given).

    With this setup, what is the potential at a point 10cm in from one plate and 10cm off the axis?

    (You can use various equations to show that the units of N/C are equivalent to V/m, both being the measure of electric field strength.)
  4. Sep 6, 2012 #3
    It's fine to find your charge first, but it's not strictly necessary, you can continue to manipulate equations until you're ready to solve for the final answer (which in almost every case saves you time).

    That said, you're on the right track, didn't check everything, but I noticed it seems you're using two different values for r. When you solved for q you used r=0.1 and when you solved for V you used r= sqrt(200).
  5. Sep 7, 2012 #4
    Hi NascentOxygen - I talked to my professor today and he said the potential at that point (10 cm, 10 cm) is 5.5*10^3 N/C because it lies in the same direction as the electric field.

    He also suggest I use: V = Ed

    V = 5.5*10^3 N/c (0.10 m) = -550 V

    I forgot why he told me the answer is negative, though. I don't seem to understand that part, especially since it is in the same direction as the electric potential.
  6. Sep 7, 2012 #5
    You can sub in q = (E*r^2)/k into v = kq/r to get V=Er but even simpler is to just look at the units. You're asked for V which is joules per coulomb. You're given N/C and a distance. How can you make N/C into J/C ? A joule is simply a newton x distance (recall work = force * distance for example). So now you know instinctively so to speak that your equation is V=Er. That is to say J/C = N/C * distance.

    So what about the positive and negative, how can you tell? Well ask yourself what is a potential difference? It's the energy required to move that charge from point A to point B. Which is point A? The one you start at. How do you know the origin is point A? Because it says V=0 at (0,0).

    Think of it in terms of gravity. Water at the top of a mountain has lots of stored potential energy... it may fall through a turbine and do work for us. In terms of this question, the charge "falls" through the electric field from A to B, and as it falls, it loses its potential energy. So use a negative sign. When charges "fall" though a circuit from a high potential to a low potential, they power our electronics :).

    With the field -> loss of potential energy.
    Against the field -> gaining potential energy.
    Last edited: Sep 7, 2012
  7. Sep 8, 2012 #6


    User Avatar

    Staff: Mentor

    The direction of an electric field is the direction that a unit positive charge would move under the influence of that field. Being positively charged, it moves away from the positive electrode towards the negative. Use this information, together with the details you are given for the direction of the field, and the voltage, to determine polarity.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook