# About the electric field inside a capacitor

If I have a metal sphere (a conductor) of radius $$R$$, carrying charge $$Q_1$$ with potential $$V_1$$, surrounded by a thick concentric metal shell (a conductor with inner radius $$a$$, outer radius $$b$$, as in Figure), which carries, a charge $$Q_2$$ on is outer surface, and $$Q_{2i}$$ on is inner surface. The shell has potential $$V_2$$. I would like to understand why the field inside depends only on $$Q_1$$ and not on $$Q_2$$ ? In other terms, why = $$V_1-V_2=\int_{1}^{2}\vec{E}\cdot \vec{dl}$$ where $$E=\frac{Q_1}{4\pi\epsilon_0 r^2}$$ where $$E$$ is the magnitude of $$\vec{E}$$. Any help would be appreciated,

• Use Gauss's Law – SK Dash Dec 8 '19 at 10:55
• So I have $\int_{\Sigma}\vec{E}\cdot \vec{d\sigma}=\frac{Q_{enclosed}}{\epsilon_0}$, this means that $\vec{E}$ is only due to $Q_1$ ... ? $\Sigma$ is a surface between the two conductors. – Dicordi Dec 8 '19 at 11:00

If you try to apply Gauss's law to a sphere concenteric with the shells and having a radius $$r$$ such that $$c(where $$c$$ is the radius of the charged sphere and $$a$$ is the inner radius of the charged shell), then you'd get,

$$\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_1}{\epsilon_0} \tag{1}$$

You can also see that the field is radially symmetric due to the symmetry of the charge distribution and also it always points in the radial direction. So,

$$\mathbf{E} \cdot d\mathbf{S} = |\mathbf{E}| |d\mathbf{S}|cos(0)= |\mathbf{E}| |d\mathbf{S}|$$

Also since $$|\mathbf{E}|$$ is constant, so,

$$\oint_S |\mathbf{E}||d\mathbf{S}|= |\mathbf{E}|\oint_S|d\mathbf{S}|= |\mathbf{E}|×4\pi r^2$$

Therefore,

$$|\mathbf{E}|=\frac{Q_1}{4\pi\epsilon_0 r^2} \tag{2}$$

As you can clearly see that the magnitude and the direction of the electric field ($$\mathbf{E}$$) only depends on $$Q_1$$ and $$r$$. The direction of the electric field can reverse if the sign of the charge $$Q_1$$ is reversed. To be precise,

$$\mathbf{E}=\frac{Q_1}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}$$

where $$\hat{\mathbf{r}}$$ is the unit vector pointing radially outwards.

Caution :- The reason why I did not conclude that the electric field($$\mathbf{E}$$) is independent of other charges except $$Q_1$$ directly from $$(1)$$ is because it only suggests that $$\int_S \mathbf{E} \cdot d\mathbf{S}$$ is independent of $$Q_1$$. You cannot conclude anything more about the electric field only by using $$(1)$$. You need to invoke symmetry arguments to prove the independence of the electric field with respect to $$Q_2$$ and $$Q_{2i}$$.

Imagine that we first charge the metal shell up to a positive charge $$Q_2'$$ by drawing electrons from it. The remaining electrons (against the background of nuclei) will redistribute themselves by Coulomb forces such that a minimum magnitude of charge density (minimum energy) is achieved. This means that all the positive charges will be distributed on the exterior surface of the shell. The charge distribution inside will be zero everywhere. Next we charge up the inner sphere up to positive charge $$Q_1$$. Here in a similar fashion, the charges will redistribute themselves on the surface of the sphere to create the minimum magnitude of charge density (which happens when all charges are distributed on the outer surface of the sphere). The positive charges $$Q_1$$, by Coulomb forces will attract negative charges from the shell. As a result, a negative surface charge, equal in magnitude to $$Q_1$$ but opposite in sign will form on the inner surface of the shell. Thus $$Q_{2i}=-Q_1$$ and the charge on the outer surface of the shell will be $$Q'_2+Q_1$$. The outer surface of the inner sphere and the inner surface of the shell form a spherical capacitor whose field can be easily calculate. Thus, this field is only determined by $$Q_1$$.
We arrive at the same conclusion if we look at any point inside at some radius $$r$$ greater than the radius but smaller than the inner radius of the shell we note that the total charge inside the sphere defined by $$r$$ is $$Q_1$$ by Gauss' Theorem this charge determines the field up to the inner radius of the shell. Inside the shell the field is of course $$0$$ since the totals charge enclosed in a similar Gaussian sphere is zero.

• This is a good answer. The suggestions that Gauss' Law answers the question are not correct. Gauss' Law gives the right answer to a calculation, but does not answer the OP's question "I would like to understand ...". However, there is a weak point here. Can you clarify why $Q_{2i}=-Q_1$? – garyp Dec 8 '19 at 14:19
• This follows from the last sentence. As I explained charge distributed by Coulomb forces and stays only on the exterior surfaces of the shell. Since charge density is zero inside the conductor the field is also zero (we can take small Gaussian surface to show that). Now If we consider a spherical concentric sphere whose radius is between the inner and outer radii of the shell then the normal field would be zero and from Gauss Theorem also the total inside charge. Thus $Q_{2i}+Q_1=0$. – am301 Dec 8 '19 at 14:42
• Yes, but this argument depends on Gauss' Law. I like the argument in the first paragraph because it does not depend on Gauss's Law. If you can support the conjecture that the charges on the facing surfaces are equal in magnitude without invoking Gauss' Law, the argument would be more attractive. – garyp Dec 8 '19 at 16:10
• Garyp, I kind of understand the all process, BUT, your two conductors are charged positively ? My question : does the charge configuration at the end depend on the charge of the two conductors ? If we charge the metal shell negatively, then, would the distribution of the charges on the metal shell be the same as your case (positively charged), if we consider that the charge of the inside sphere is positive ? – Dicordi Dec 8 '19 at 16:16
• garyp - Gauss' Law is pretty fundamental in electrostatics so I am not sure we can without it. It is equivalent to Coulomb's Law that is at the basis of electrostatics. Perhaps we can also think about each charge on the inner sphere generating a radial field line that (must be) perpendicular to the surface. Such field lines terminates at the inner surface of the shell at an opposite charge and hence the numbers of charges on the surfaces must be equal. – am301 Dec 8 '19 at 19:24

Since electromagnetism obeys the superposition principle the problem can be decomposed. It is clear that the inner sphere contributes $$E_1=\frac{Q_1}{4\pi\epsilon_0 r^2}$$ to the field in between the spheres. Because of superposition all you need to establish is the field of the empty sphere 2, without sphere 1 inside. By Gauss's law you know that the static field of a charged surface enclosing an empty volume is zero inside that volume. The end result is that only sphere 1 determines the field.

The physical reason is that electrostatic potential on a conductor is constant. Therefore no E field lines can connect any two points on the conductor as on the inside of the surface any field line must land back on the surface. Loops are impossible as the electrostatic potential is conservative. On the outside of the sphere a field line can extend to infinity.