thanks. I didn't think about it this way but it is definitely clear after your comments.

Thanks for your help. I see that my initial assumption was correct and I did misunderstand your remark. I will pass this onto my friend.

great link! At this level (first course in EM), I think that one is not asked to consider such questions. I am a third year student and I find this article really interesting and thought provoking.

thanks for the response. The question is about the field at the surface of the sphere. As I get really close to the surface, the field behaves as an infinite plane with field $E = σ/ϵ_0$, so it appears that surface charge density is the answer. I agree with you that if I was away from the surface then $E_2$ would be the smaller field.

@JohnRennie: sorry to say but I don't think that I understand your comment. If I do this, I believe that I find that the E-fields are equal. That is, $F_1∝q_1E_1$ then $F_2/F_1=(q_2^2/r_2^2)/(q_1^2/r_1^2)=1$. This doesn't make any sense.

@rijulgupta: Why is the electric field $E_2 > E_1$? What's the physics that explains why $E_2 > E_1$? I want to say that the surface charge density is greater for sphere 2 than sphere 1 since sphere 1 is larger than sphere 2. But I am not sure.

@Carlos: You have a typo in your answer. The polarization and E-field relation in the first paragraph is reversed. Really nice answer, more physical than the book's explanation.