1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Fields- 2 Dimensions

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-57 due to the two positive charges (Q = 4.0 µC) shown.


    3. The attempt at a solution

    Basically, I'm completely lost. I've applied Coulomb's law to find the magnitudes of the contributing electric fields.

    Can anyone help me solve this problem/ at least get started. Thanks so much, god bless.
  2. jcsd
  3. Jan 24, 2009 #2
    Apply Coulomb's law for each charge separately.
    Then use vector superposition once you find electric field for each charge.
  4. Jan 25, 2009 #3
    An electric field is a vector so it has a magnitude and a direction. Using Coulomb's Law gives an expression for E:

    [tex]\vec{E}=\frac{\vec{F}}{q}\mbox{ where q is a test charge}[/tex]

    [tex]\vec{F}=\frac{qQ\hat{r}}{4\pi \varepsilon_0 \ r^2}[/tex]

    [tex]\mbox{where }\hat{r}\mbox{ is the unit vector in the r direction}[/tex]


    [tex]\vec{E}=\frac{Q\hat{r}}{4\pi \varepsilon_0 \ r^2}[/tex]

    where r is the distance from the charge to the point in question. Resolve E into Ex and Ey using cosine and sine. Then add these resolved components to find the resultant components.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric Fields- 2 Dimensions
  1. Electric fields 2 (Replies: 5)