I'm trying to do a past exam paper on electromagnetism for communications. One of the questions is as follows:

You are required to find the charge enclosed within the surface $\partial V$ of a sphere $V$ of radius $R$ centered at the origin of coordinates $(0, 0, 0)T$ due to an electrostatic field described by the equation

$E = i (x^3 + 2x(y)^2 + 2x(z)^2 > +j ((y)^3 + 2y(z)^2 + 2y(x)^2) +k ((z)^3 + 2z(x)^2 + 2z(y)^2)$

  • Which of Maxwell’s equations is needed for this task, and how is it used?
  • How can you verify that there are no time dependent magnetic fields present in this situation?

You will find the parameterisation $\alpha : (r, \theta, \varphi) \to (x, y, z)$ of the sphere $V x^2 + y^2 + z^2 \leq R_2$ in terms of spherical polar coordinates $(r, \theta, \varphi)$ useful, where $\theta$ and $\varphi$ are the co-latitudinal (polar) and longitudinal (azimuthal) angles:

$x = r \sin \theta \cos \varphi$

$y = r \sin \theta \sin \varphi$

$z = r \cos \theta$,

and $r = R$ on the surface. You will find that using $\int\partial V \omega = \int Vd\omega$

duly specialized for the appropriate differential forms $\omega$ simplifies the calculations.

This is what i think i need to do:

-We use Gauss Law

-We know by definition for Gauss law $\int\partial V \ E dA = \delta (dot product) E = Q/e0 $

-$ surface area of a sphere is dA = r^2sin(\theta)d\theta d\varphi$

ok so find div of E, then prove its electrostatic with curl E = 0,

with the div E we do a surface integral for a sphere replacing the remaining x,y,z with the corresponding r phi and theta, some cos and sin will cancel out and after surface integral and replacing r = R i have E = 4e0(pi)(R)^4 this is the enclosed electric field = enclosed charge by definition of Gauss law

is this the right method? I'm getting confused with the surface integral part.


closed as off-topic by Kyle Oman, sammy gerbil, Jon Custer, John Rennie, Kyle Kanos Jan 6 '17 at 11:09

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Use Gauß's law $\iiint\nabla.\vec{E}\mathrm{d}V=\frac{q}{\epsilon}$

Here $\mathrm{d}V=r^2\sin{\theta}\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$


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