1. Limited time only! Sign up for a free 30min personal 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!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    Way to go, Reshma!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook