Gauss's Law Physics Question?

Question

A solid sphere of radius 40.0 cm has a total positive charge of 32.0 µC uniformly distributed throughout its volume. Calculate the magnitude of the electric field at the following distances from the center of the sphere.
(a) 0 cm (b) 10.0 cm (c) 40.0 cm (d) 60.0 cm
units are in kN/C

i've tried to use the formula E=ke(Q)/(r2) but seems like not the right formula. anything will be helpful! thanks

i did tried to use the formula. but appear to not get the right answer still. For Q=26uC, do i need to convert that? i use the equation: E= (1/4piE0)(q/r^2) is it right?

Answer 1

Gauss' law says that the integral of (vector)E•dS over a closed surface S equals a constant times the charge enclosed by S. (If you're using MKS units, the constant is 1 divided by ε0, the permittivity of free space.) Consider S as a sphere of radius r, with the same center as that of the solid sphere. By symmetry, (vector)E is everywhere normal to the surface and constant in magnitude over the surface, so the surface integral is (4πr²)E where E is the magnitude of (vector)E.

If the total charge enclosed by S is denoted q, then

E = kq/r²

(In MKS, k = 1/(4π ε0))

If S has radius less than that of the solid sphere, it encloses only a portion of the charge contained in the solid sphere, and only that enclosed charge contributes to E; the contributions from the charges outside S cancel out. So E at the center of the solid sphere is zero. (This is also clear from symmetry.)

And if you are using MKS units, don't forget to change the distances in (b)-(d) to meters.