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Electric Field: Continuous Charge Distribution

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A nonconducting sphere 1.3 m in diameter with its center on the x axis at x = 4 m carries a uniform volume charge of density ρ = 4.8 µC/m3. Surrounding the sphere is a spherical shell with a diameter of 2.6 m and a uniform surface charge density σ = -1.2 µC/m2. Calculate the magnitude and direction of the electric field at the following locations.
    1) x = 4m, y = 1.2m
    2) x = 2m, y = 3m
    Both answers are in components of i[hat] and j[hat] vectors with units of N/C.

    2. Relevant equations
    Well, the textbook solutions use the following equations...
    Eshell = kQshell/r2 r[hat]
    It uses three different equations for the sphere for each part. Not sure where it got most of these equations as in the textbook itself, these are the equations. I can include the solutions' equations but when I tried their way of solving, my answer was wrong.
    Esphere = Qk/r2 (when r is greater than or equal to R)
    Esphere = Qkr/R3 (when r is less than or equal to R)

    3. The attempt at a solution
    I know that the x component of 1) is 0 because that's where the center of the sphere is at. However I'm honestly stumped on how to solve for the y component of 1) and subsequently how to solve 2) in general. As I already mentioned I tried the equations in the textbook solutions but they gave me the wrong answer and didn't explain why they used certain numbers or equations.

    For 1), the electric field of the shell is 0 since the point is within the shell. I tried the equations in the textbook itself and I guess (and am absolutely confident this is wrong) the equation for the sphere's electric field is

    Esphere = k*ρ*4πr4/R3 r[hat]

    I have limited tries on the online homework so detailed help would be greatly appreciated. I tried reading the textbook as well as other textbooks and websites and online videos and I'm still stuck. Thanks.
     
  2. jcsd
  3. Oct 18, 2015 #2

    gneill

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    Staff: Mentor

    If you draw a vector from the center of the spheres to a given point, how would you find the x and y components of that vector (think right triangles)? How do you form a unit vector from a given vector?

    Layout.png
     
  4. Oct 18, 2015 #3
    The x and y components would be the radius (which is the hypotenuse, if I'm not mistaken) times cos θ and sin θ respectively. And the unit vector of a given vector is the x and y components of that vector divided by the magnitude of the vector itself.
     
  5. Oct 18, 2015 #4

    gneill

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    Sure. But you are given the actual points so you have a way to determine the components directly. Here's the same image with one of the vectors drawn:
    Layout2.png

    You should be able to see the vector components. Use the given point location coordinates to calculate the components.
     
  6. Oct 18, 2015 #5
    So I got that the magnitude of the vector is approximately 3.606m at an approximate angle of 56.309°.

    ##\vec{r} = -3.606 \cos(56.309) \hat{i} + 3.606 \sin(56.309) \hat{j}##

    Thus ##\hat{r} = -0.5547 \hat{i} + 0.83205 \hat{j}##.

    Would I use that ##\hat{r}## for the equation of the sphere's electric field?
     
  7. Oct 18, 2015 #6

    gneill

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    You've done a lot of extra work using trig functions. You don't need to find the angle. You have a right angle triangle whose sides ARE the components of the vector. The starting and ending points of the vector are all you need: Δx and Δy.

    You use the magnitude of the actual vector in the field magnitude part of the expression, the unit vector (##\hat{r}##) just gives the directional component. Since it has unit magnitude it doesn't change the magnitude of the result.
     
  8. Oct 18, 2015 #7
    So would that mean that I only use the fact that ##\hat{r} = - \hat{i} + \hat{j}##?

    Also, if my equation for the electric field of the sphere is correct, would I plug in r = 3.606m and R = 1.2m?
     
  9. Oct 18, 2015 #8

    gneill

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    No, the unit vector here is a vector of unit length that points in the direction of the vector in question. ##- \hat{i} + \hat{j}## does not have a magnitude of unity, and it doesn't point in the direction of either of the vectors of interest in this problem.

    The use of unit vectors is a way to separate the magnitude and direction parts of a given vector. Say that you have some vector ##\vec{R}##. It has a magnitude ##R = |\vec{R}|##. Now, if ##\vec{r}## is unit vector in the same direction as ##\vec{R}##, then you can write:

    ##\vec{R} = R \vec{r}##

    You can then use scalar operations to handle the magnitude calculations (avoiding all that slogging through vector dot and cross products), and then tack on the correct unit vector to give the direction.

    You found the unit vector for the case under discussion as ##\hat{r}## in post #5.
    I'm not sure if I'm interpreting your meaning correctly. Can you be more explicit? How about showing the full calculation for the point at (2,3) m ?
     
  10. Oct 18, 2015 #9
    I have an attempted equation (excluding values) for the electric field of the sphere toward the end of my original post,

    Esphere
    = k*ρ*4π*r4/R3 r[hat]

    I don't know how to do the full calculation at point (2m, 3m). but the above equation is my guess of what to use to find the electric field of the sphere at the point (4m, 1.2m)
     
  11. Oct 18, 2015 #10

    gneill

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    Okay, that doesn't look right to me. Perhaps the meaning of r and R are changing as we go along? There are a lot of different radii in this problem.

    Can you spell out the definitions of various r's in that equation? I'm thinking that I should be able to spot the calculation for the volume of the inner sphere in there (for calculating the total charge of the sphere), but I'm not seeing it. Perhaps you should detail your derivation. Start with defining the variables and then proceed to calculate the charge on the sphere.
     
  12. Oct 18, 2015 #11
    Here is a picture from my textbook that would probably illustrate what the differences between r and R are better than I could by explaining it. I used the equation of the electric field when r is less than or equal to R and set Q = ρV. Since the volume of a sphere is 4πr3, I substituted all of those things and got that really crazy equation.

    When I plug in my numbers (with r = 0.65 m and R = 1.2m) I obtain 56004.61984 N/C but that doesn't seem correct.
     

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  13. Oct 18, 2015 #12

    gneill

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    Check your formula for the volume of a sphere. You're missing a constant.

    The point you are dealing with is outside the sphere, so r > R. The spherical charge should "look like" a point charge to the observer at the point in question, so the usual Coulomb's law version for point charges will work fine there.

    You can check the sanity of an equation by making sure that the units balance and work out to what you expect. Plug in unis for the variables and see if they boil down to what you expect.
     
  14. Oct 19, 2015 #13
    Ahhh I forgot to divide by 3. I checked the variables and the units should come out to N/C if I am working on the electric field of the sphere at (4m, 1.2m).

    Are you referencing the point (2m, 3m) as being outside of the sphere?
     
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