Homework question about electric field between two spheres I was helping a friend of mine with the following question from Knight's book and I was not able to answer part (c).



*The two metal spheres in FIGURE Q30.9 are connected by a metal wire with a switch in the middle. Initially the switch is open. Sphere 1, with the larger radius, is given a positive charge. Sphere 2, with the smaller radius, is neutral. Then the switch is closed. Afterward, sphere 1 has charge $Q_1$ is at potential $V_1$, and the electric field strength at its surface is $E_1$. The values for sphere 2 are $Q_2$, $V_2$, and $E_2$.

a. Is $V_1$ larger than, smaller than, or equal to $V_2$? Explain.
b. Is $Q_1$ larger than, smaller than, or equal to $Q_2$? Explain.
c. Is $E_1$ larger than, smaller than, or equal to $E_2$? Explain.

FIGURE Q30.9

Here is what I think I know:
(a) I expect $V_1 = V_2$; the two spheres are in equilibrium so no current flows between the two spheres.
(b) If $V_1 = V_2$, then $Q_1/R_1 = Q_2/R_2$. This implies that $Q_1 = Q_2 (R_1/R_2) > Q_2$. That is, $Q_1$ is larger than $Q_2$.
(c) Here is were I am unsure: I think that $E_2 > E_1$.
Question: Mathematically it makes since because if $V_1 = V_2$, then $E_1 = V_1/R_1$ and $E_2 = V_2/R_2$, then $E_2 > E_1$. But I don't see this physically. Can someone explain this?
 A: The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is:
$$ F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2} $$
and the field strength at the surface of the second sphere is:
$$ F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2} $$
Lets take the ratio $F_1/F_2$ to see which is greater. The constants cancel to give us:
$$ \frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}} $$
and I'm going to rewrite this slightly to make it obvious how you use your equality $Q_1/r_1 = Q_2/r_2$:
$$ \frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}} $$
Because $Q_1/r_1 = Q_2/r_2$ we can cancel them on the top and bottom of the fraction and we're left with:
$$ \frac{F_1}{F_2} = \frac{r_2}{r_1} $$
and because $r_2 < r_1$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1.
A: 
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

Don't forget that the surface area goes as the square of the radius.  As you wrote:
$$ Q_2 = Q_1 (R_2/R_1)$$
but the surface area of sphere 2 is:
$$A_2 = A_1 (R_2/R_1)^2 $$
thus:
$$\dfrac{Q_2}{A_2} =  \dfrac{Q_1}{A_1}\dfrac{R_1}{R_2} > \dfrac{Q_1}{A_1}$$
A: For physical interpretation, I understand you are confusing the field to be related to surface charge density only. But, At a sufficient distance from the sphere one can consider the charge to behave as point charge, hence when more charge is accunulated on 1st sphere, it corresponds to greater electric field. 
Note : your thought about greater surface charged density would be applicable for very very closely placed spheres as theb induction would take place and dominate over other charges here as much there  is surface charge density more would be the attractiob but at greater distances the induction is negligible and you can treat spherical charge distributions as point charges.
