Gauss' law for a cavity in an insulator

[laser]

Homework Statement

A solid sphere of radius R has uniform charge density ρ. A hole of radius R/2 is scooped out of it as shown in Figure 10. Show that the field inside the hole is uniform and along the x-axis and of magnitude ρR/6ε0. Hint: Think of the hole as a superposition of positive and negative charges.

Relevant Equations

E=kq/r^2

This is a problem from Yale OCW (Shankar). The solution he gives is as follows:

Sure, this makes sense. However...

Superimposed rho and negative rho with radius R/2 means there is no charge enclosed in the cavity... therefore
no charge -> no flux -> no electric field.

 

[Orodruin]

Homework Statement: A solid sphere of radius R has uniform charge density ρ. A hole of radius R/2 is scooped out of it as shown in Figure 10. Show that the field inside the hole is uniform and along the x-axis and of magnitude ρR/6ε0. Hint: Think of the hole as a superposition of positive and negative charges.
Relevant Equations: E=kq/r^2

This is a problem from Yale OCW (Shankar). The solution he gives is as follows:

View attachment 340207

Sure, this makes sense. However...

Superimposed rho and negative rho with radius R/2 means there is no charge enclosed in the cavity... therefore
no charge -> no flux -> no electric field.

No. You cannot draw the conclusion that there is no electric field. You would need to be able to make some symmetry argument for that to hold.

No net flux only means any field lines that come in also go out somewhere else.

 

[Orodruin]

This is also somewhat applicable https://www.physicsforums.com/insights/a-physics-misconception-with-gauss-law/

You should also provide the relevant figure when you make a post like this.