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Homework Help: Problem on electric field

  1. Dec 25, 2005 #1
    I have the electric field in a region give by:
    [tex]\vec E = \frac{k}{\epsilon_0 a^4} \left(ax^2\hat x + yz^2\hat y + y^2z\hat z\right)[/tex]
    where 'k' and 'a' are constants.
    There are few questions I need to solve.

    1: Is the field conservative?
    A: Yes. I computed the curl and found it equal to zero.
    [tex]\nabla \times \vec E = 0[/tex]

    2: Calculate the charge density at a point P(x,y,z).
    A: I applied the differential form of Gauss's law.
    [tex]\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}[/tex]
    [itex]\rho[/itex] is the charge density.
    So, I got the charge density as:
    [tex]\rho = \epsilon_0 \left[2ax + y^2 + z^2\right][/tex]
    Please verify if my method is correct.

    3: Determine the charge enclosed in a cube of side 'a' with one of its corners at the origin and sides parallel to x,y and z axes.
    A: I applied Gauss's law here too!
    [tex]\int \vec E \cdot d\vec a = \frac{Q_{enclosed}}{\epsilon_0}[/tex]
    How do I calculate the flux here?
    Last edited: Dec 25, 2005
  2. jcsd
  3. Dec 25, 2005 #2

    Tom Mattson

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    Your method is correct.

    Right again.

    Calculate it over each face of the cube, one at a time. Here's a hint: For the two faces that are parallel to the yz-plane, the unit normals are [itex]\hat{i}[/itex] and [itex]-\hat{i}[/itex].
  4. Dec 25, 2005 #3


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    Problem 3 is ambiguous since any one of the 8 corners of the cube could be at the coordinate origin - which is probably a hint at what the answer will be! :)

    All you need to do is recognize that [itex]d\vec a[/itex] is in the direction of the normal to the surface and consists of products of dxdy, dxdz and dydz - then integrate over all 6 faces of the cube.
  5. Dec 27, 2005 #4
    Thanks Tom Mattson and Tide for looking into my problem!
    Er..actually I made a serious blunder here :uhh: .
    I forgot to multiply the constant factor in the electric field expression. So the charge density is:
    [tex]\rho = \frac{k}{\epsilon_0 a^4} \epsilon_0 \left[2ax + y^2 + z^2\right][/tex]
    [tex]\rho = \frac{k}{a^4} \left[2ax + y^2 + z^2\right][/tex]
  6. Dec 27, 2005 #5
    Thanks, I tried it. There are 6 faces in all.
    1] x = a, [itex]d\vec a = dydz\hat x[/itex]

    [tex]\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a a^2 dydz = \frac{ak}{\epsilon_0}[/tex]

    2] x = 0, [itex]d\vec a = -dydz\hat x[/itex]
    Intergal vanishes here!

    3] y = a, [itex]d\vec a = dxdz\hat y[/itex]

    [tex]\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a az^2 dxdz = \frac{ak}{3\epsilon_0}[/tex]

    4] y = 0, [itex]d\vec a = -dxdz\hat y[/itex]
    Integral vanishes here too.

    5] z = a, [itex]d\vec a = dxdy\hat z[/itex]

    [tex]\frac{k}{\epsilon_0 a^4} \int_0^a \int_0^a ay^2 dxdz = \frac{ak}{3\epsilon_0}[/tex]

    6] z = 0, [itex]d\vec a = -dxdy\hat z[/itex]
    Integral vanishes here.

    So the total flux is:
    [tex]\frac{ak}{\epsilon_0} + \frac{ak}{3\epsilon_0} + \frac{ak}{3\epsilon_0} = \frac{5ak}{3\epsilon_0}[/tex]

    Hence, total charge enclosed is:
    [tex]Q_{enclosed} = \epsilon_0 \frac{5ak}{3\epsilon_0} = \frac{5ak}{3}[/tex]
    Last edited: Dec 27, 2005
  7. Dec 27, 2005 #6
    An alternative method

    I discovered yesterday that there is yet another alternative method to solve this problem. I can use the Green's theorem over a volume element.
    [tex]Q_{enclosed} = \int_{all space} \rho dV[/tex]
    dV is the volume element given by [itex]dV = dxdydz[/itex]
    [tex]Q_{enclosed} = \frac{k}{a^4}\int_0^a \int_0^a \int_0^a \left[2ax + y^2 + z^2\right]dx dy dz[/tex]
    The evaluation of this integral gives me the same result as above.
    [tex]Q_{enclosed} = \frac{5ak}{3}[/tex]
    So my answer is correct!! :biggrin:
  8. Dec 27, 2005 #7


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    Way to go, Reshma!
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