Question ca98e

May 2, 2017

Gauss Law:

$\Phi = \int {\int}_{S} m a t h b f E \cdot d m a t h b f S = \frac{\sum Q}{\epsilon} _ o$

The total flux through an enclosing surface is equal to the charge enclosed divided by a constant.

In this case, the charge is at the centre of the cube so symmetry is our friend.

The flux through one side will be the one sixth the total flux. We don't need to know the dimensions.

As long as we use a cubic Gaussian surface that contains the charge at its centre, we know that:

${\Phi}_{T o t a l} = \frac{9.6 \times {10}^{- 6}}{\epsilon} _ o$

Phi_("One side") = (9.6 xx 10^(-6))/(6 epsilon_o) approx 1.8 x 10^5 " Vm"#