Question

Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electric field falls off as 1/r2 of the distance r from the charge. How does the surface area change with r ?

Answer 1

The change in surface area of Gaussian surface with radius (r) is 8πr.

Electric field from Coulomb's law

The electric field experienced by a charge is calculated as follows;

$E = \frac{Q}{4\pi \varepsilon_o r^2}$

where;

• E is the electric field
• Q is the charge
• r is the radius

The electric field reduces by a factor of $\frac{1}{r^2}$

Surface area of a Gaussian surface;

The surface area of a sphere is given as;

$A = 4\pi r^2$

Change in area with r

$\frac{dA}{dr} = 8\pi r$

Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.

Learn more about area of Gaussian surfaces here: brainly.com/question/17060446