3
$\begingroup$

I was helping a friend of mine with the following question from Knight's book and I was not able to answer part (c).

  1. 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.

enter image description here
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?

$\endgroup$
  • $\begingroup$ What do you mean by not seeing this physcally ? $\endgroup$ – Rijul Gupta Oct 25 '13 at 9:21
  • $\begingroup$ The field strength is the force on a charge of 1 Coulomb, and this is given by the usual law for the force between two charges. If you write down the force on your test charge for a charge $Q_1$ at a distance $r_1$, do the same for sphere 2, then use the expression $Q_1/r_1 = Q_2/r_2$ you can calculate the ratio of $F_1$ to $F_2$. $\endgroup$ – John Rennie Oct 25 '13 at 9:55
  • $\begingroup$ @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. $\endgroup$ – Juan Oct 25 '13 at 10:30
  • $\begingroup$ @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. $\endgroup$ – Juan Oct 25 '13 at 10:44
  • $\begingroup$ @Juan: I've added an answer with the details $\endgroup$ – John Rennie Oct 25 '13 at 12:24
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Juan Oct 25 '13 at 13:23
2
$\begingroup$

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:

$$\dfrac{Q_2}{A_2} = \dfrac{Q_1}{A_1}\dfrac{R_1}{R_2} > \dfrac{Q_1}{A_1}$$

$\endgroup$
  • $\begingroup$ thanks. I didn't think about it this way but it is definitely clear after your comments. $\endgroup$ – Juan Oct 25 '13 at 13:24
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Juan Oct 25 '13 at 11:03
  • $\begingroup$ I would advice you neither too get so close to a conducting charged sphere and nor too expect such a simple field formula from it due to as I said its induction's realm at distances near surface. For more information check this out : nature.com/news/like-attracts-like-1.10698 $\endgroup$ – Rijul Gupta Oct 25 '13 at 11:11
  • $\begingroup$ Moreover for solving simple problems we assume that our greater distance approximations are valid upto the surface of conductors. $\endgroup$ – Rijul Gupta Oct 25 '13 at 11:19
  • $\begingroup$ 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. $\endgroup$ – Juan Oct 25 '13 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.