Aspherically symmetric charge distribution produces the electric field vector e = (100/r)r̂ n/c, where r is in m. (a) what is the electric field strength at r = 9 cm? 1111.111111 correct: your answer is correct. n/c (b) what is the electric flux through a 18-cm-diameter spherical surface that is concentric with the charge distribution?

(a) The magnitude of the electric field is given by:

where r is the distance from the centre in metres. Therefore, if we substitute

we can find the magnitude of the electric field at a distance of 9 cm:

(b) The electric field direction is radial at every point of the space (in fact, the problem gives us the direction of the electric field vector: e = (100/r)r̂ n/c, where r̂ indicates that the field is radial). Therefore, the electric field is perpendicular to the surface of the concentric sphere of radius r=18 cm=0.18 m, so the flux of the electric field through the surface of the sphere is simply equal to the product between the electric field strength calculated at the surface of the sphere and the surface of the sphere:

real, smaller and upside-down.   so the answer should be b.