The flux of the vector field \mathbf{G}(\mathbf{r}) = z^2 \hat{\mathbf{e}}_z through the surface of a sphere of radius R , centred around the origin, is equal to

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We are asked to compute the flux of G(r) = z^2 e_z through the closed surface of a sphere of radius R centered at the origin.

A systematic way is to use the divergence theorem, which relates the surface flux to a volume integral of the divergence: ∬_S G · dS = ∭_V (∇ · G) dV.

First, compute the divergence of G. Since G = (0, 0, z^2), we have ∇ · G = ∂G_x/∂x + ∂G_y/∂y + ∂G_z/∂z = 0 + 0 + ∂(z^2)/∂z = 2z.

Thus, the flux equals ∭_V 2z dV over the solid sphere of radius R.

Now consider the symmetry: the region V (the ball) is symmetric under z → −z. The integrand 2z is an odd function of z, while the domain is symmetric about the xy-plane. Therefore the integral of 2z over V cancels out and equals zero:
∭_V 2z dV = 0.

Equivalently, one can reason that the contribution to the flux from the upper hemisphere is exactly canceled by the contribution from the lower hemisphere due to the z-dependence of the vector field component being proportional to z^2 but the flux depends linearly on z through the divergence, yielding a net zero when integrated over a symmetric region.

Hence, the flux of G through the spherical surface is zero.

PHAS0009_24-25 3.3 Quiz 3 [0.25 hrs]

The flux of the vector field \mathbf{G}(\mathbf{r}) = z^2 \hat{\mathbf{e}}_z through the surface of a sphere of radius R , centred around the origin, is equal to

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