Image text transcribed for accessibility: Find the magnitude of the electric field at a point inside the sphere. In this case, let's choose a spherical Gaussian surface having radius r < a, concentric with the insulating sphere (see Figure (b)). Let V' be the volume of this smaller sphere. To apply Gauss's law in this situation, recognize that the charge qis within the Gaussian surface of volume V' is less than Q. Calculate qin by using qin = pV': Apply Gauss's law in the region r < a: Substitute p = Q/((4/3) pi a3) and epsilon 0 = Simplify E. (Use the following as necessary: ke, Q, a, and r.) This result for E differs from the one obtained in part (A). It shows that E rightarrow 0 as r rightarrow 0. Therefore, the result eliminates the problem that would exist at r = 0 if E varied as 1/r2 inside the sphere as it does outside the sphere. That is, if E 1/r2 for r < a, the field would be infinite at r = 0, which is physically impossible.