Potentials (and ##\Delta\phi##) of nearby converging bodies

[sergiokapone]

Homework Statement

An uncharged insulated conductor is brought to a positively charged solitary conductor. Show that the potentials of both conductors will increase, and the potential difference between them decreases.

Relevant Equations

\begin{equation}
\phi_i = \sum D_{ij} Q_j.
\end{equation}
$D_{ij}$ - voltage matrix.

Let $Q$ - charge of one of conductor, $\phi_1$ --- potential of charged conductor, $\phi_2$ --- potential of uncharged conductor.
For the charged conductor:
\begin{equation}
\phi_1 = D_{11}Q ,
\end{equation}

for uncharged conductor:
\begin{equation}
\phi_2 = D_{21}Q
\end{equation}

 

[TSny]

Problem Statement: An uncharged insulated conductor is brought to a positively charged solitary conductor.

Can you please clarify the phrase "brought to"? Does it mean that the uncharged conductor is brought near to the charged conductor, but they do not touch?

Show that the potentials of both conductors will increase

I don't think the potentials of both conductors will increase. Maybe I'm misinterpreting the problem. Is this a problem from a standard textbook? Have you stated the problem exactly as given to you?

 

[sergiokapone]

Can you please clarify the phrase "brought to"? Does it mean that the uncharged conductor is brought near to the charged conductor, but they do not touch?

May be, this is wrong translation from russian. Pisiible it is better to say "getting closer". Yes, they do not touch.

Is this a problem from a standard textbook? Have you stated the problem exactly as given to you?

This problem from russian book by Sivukhin "General physics". Vol 3. Electricity.

This is not my homework. I just interested in this problem, because I think is wrong.

 

[TSny]

May be, this is wrong translation from russian. Pisiible it is better to say "getting closer". Yes, they do not touch.

This problem from russian book by Sivukhin "General physics". Vol 3. Electricity.

OK

This is not my homework. I just interested in this problem, because I think is wrong.

Yes, I think the potential of the uncharged conductor will increase while the potential of the charged conductor will decrease. The potential difference between the two conductors will decrease.

 

[sergiokapone]

potential of the charged conductor will decrease.

I think, the potential of charged conductor will remain unchanged. I conclude this from consideration of special case of conductors - spherical conductors.

 

[TSny]

I think, the potential of charged conductor will remain unchanged. I conclude this from consideration of special case of conductors - spherical conductors.

The figure below shows the charged conductor when the uncharged conductor is far away. The potential of the charged conductor can be obtained by integrating E from point A out to infinity.

The figure below shows (roughly) the situation when the uncharged conductor is nearby. Consider the magnitude of the E field along the horizontal field line marked with A. How does it compare with the previous figure? How does the potential at A here compare with the potential at A in the figure above?

 

[sergiokapone]

I reasoned in terms of potentials.
Consider spheres with radii $R_1$ and $R_2$ at a distance $\ell > R_1 + R_2$ and write appropriate equations:

For the first (charged) sphere
\begin{equation}
\phi_1 = \frac{1}{R_1} q_1 + \frac{1}{\ell} q_2 ,
\end{equation}

for the uncharged sphere
\begin{equation}
\phi_2 = \frac{1}{\ell} q_1 + \frac{1}{R_2} q_2.
\end{equation}Since the sphere is not charged ($q_2 = 0$), we get
\begin{equation}
\phi_1 = \frac{1}{R_1} q_1 ,
\end{equation}

\begin{equation}
\phi_2 = \frac{1}{\ell} q_1.
\end{equation}

So, we can see, the potential of fist sphere does not change with decreasing $\ell$ ($\ell \to R_1 + R_2$), and the potential of uncharged sphere does increasing. Potential difference $\phi_1 - \phi_2$ does decreasing.

 

[sergiokapone]

Consider the magnitude of the E field along the horizontal field line marked with A. How does it compare with the previous figure? How does the potential at A here compare with the potential at A in the figure above?

From consideration of your pictures, I can't conclude anything about spheres potentials. But, I see, the field between spheres become much stronger, so I can conclude, the potential difference become smaller.

 

[TSny]

I reasoned in terms of potentials.
Consider spheres with radii $R_1$ and $R_2$ at a distance $\ell > R_1 + R_2$ and write appropriate equations:

For the first (charged) sphere
\begin{equation}
\phi_1 = \frac{1}{R_1} q_1 + \frac{1}{\ell} q_2 ,
\end{equation}

I don't see how you get this. When sphere 2 is near sphere 1, the surface charge density on sphere 1will be distributed non-uniformly over the surface of the sphere. As a result, the potential at the surface of 1 due to the charge on 1 will not be $q_1/R_1$. Also, the potential at the surface of 1 due to the charge $q_2$ will not simply be $q_2/ \ell$.

 

[sergiokapone]

I don't see how you get this. When sphere 2 is near sphere 1, the surface charge density on sphere 1will be distributed non-uniformly over the surface of the sphere. As a result, the potential at the surface of 1 due to the charge on 1 will not be $q_1/R_1$. Also, the potential at the surface of 1 due to the charge $q_2$ will not simply be $q_2/ \ell$.

Yes, I agree with you. My formulas is approximate and valid for case $\ell \gg R_1 + R_2$.

 

[TSny]

From consideration of your pictures, I can't conclude anything about spheres potentials. But, I see, the field between spheres become much stronger, so I can conclude, the potential difference become smaller.

Due to the rearrangement of charge on the surface of 1 when 2 is brought near, the electric field along the field line containing point A will be weaker in the second picture compared to the first picture. So, the integral of E along this line from A out to infinity will be less for the second picture. Thus, the potential at A in the second picture is less than the potential at A in the first picture.

 

[TSny]

Yes, I agree with you. My formulas is approximate and valid for case $\ell \gg R_1 + R_2$.

But you need to consider the case where $\ell$ is not much larger than $R_1+R_2$, since the question is asking about what happens when 2 is brought near to 1.

 

[sergiokapone]

the electric field along the line with point A will be weaker in the second picture compared to the first picture.

I do not understand. In diametrally oposite point, say $B$, the field is stronger (in second picture), so, we can integrate $\int\limits_{B \to \infty} \vec{E}\cdot d\vec{r}$ (through 2nd sphere).

 

[sergiokapone]

I'm thinking, maybe it is need to think about second sphere as a dipole, and consider it potential as a potenial of dipole...

 

[TSny]

I do not understand. In diametrally oposite point, say $B$, the field is stronger (in second picture), so, we can integrate $\int\limits_{B \to \infty} \vec{E}\cdot d\vec{r}$ (through 2nd sphere).

Yes, in principle you could start in the second picture at $B$ (which is diametrically opposite to $A$) and integrate $E$ along a horizontal line from $B$, through sphere 2, and on out to infinity. You would necessarily get the same result as starting at $A$ and integrating out to infinity. But, the advantage of starting at $A$ is that it is easy to compare with starting at $A$ in the first picture.

 

[sergiokapone]

But you need to consider the case where $\ell$ is not much larger than $R_1+R_2$, since the question is asking about what happens when 2 is brought near to 1.

Yes. Ok. Let's forget about this case

 

[TSny]

Yes. Ok. Let's forget about this case

I'm not sure I understand. Are you saying that we should forget about the case where sphere 2 is near sphere 1?

 

[sergiokapone]

I'm not sure I understand. Are you saying that we should forget about the case where sphere 2 is near sphere 1?

I mean, forget about case $\ell \gg R_1 + R_2$.

 

[sergiokapone]

You would necessarily get the same result as starting at $A$ and integrating out to infinity. But, the advantage of starting at $A$ is that it is easy to compare with starting at $A$ in the first picture.

Yes, integrating alohg $B \to\infty$ and $A \to\infty$ must give the same result. That is, we start from points with the same potential but with different values of the field and we get the same result. Therefore, it seems to me that the reasoning by analyzing the density of the lines of the field is not entirely strict. But, most likely, you are right.

 

[TSny]

I mean, forget about case $\ell \gg R_1 + R_2$.

OK. But we want to compare the case where $\ell \gg R_1 + R_2$ to the case where 2 is near 1. That is, compare the first picture with the second picture.

 

[sergiokapone]

I wonder, if it can be mathematically more rigorously shown that the potential of the former should decrease.

 

[sergiokapone]

OK. But we want to compare the case where $\ell \gg R_1 + R_2$ to the case where 2 is near 1. That is, compare the first picture with the second picture.

I think, if we obtain mathematical estimation for case $\ell > R_1 + R_2$ then compare with the case $\ell \gg$.

 

[TSny]

Yes, integrating alohg $B \to\infty$ and $A \to\infty$ must give the same result. That is, we start from points with the same potential but with different values of the field and we get the same result. Therefore, it seems to me that the reasoning by analyzing the density of the lines of the field is not entirely strict. But, most likely, you are right.

Let $C$ be any point on the horizontal line to the right of sphere 1. I think it is fairly intuitive that the field at $C$ in the second picture is weaker than the field at $C$ in the first picture.

Admittedly, this is not rigorous. It is possible to construct a more rigorous argument.

 

[TSny]

I wonder, if it can be mathematically more rigorously shown that the potential of the former should decrease.

Yes. One way is to use energy concepts. You can use Thomson's theorem of electrostatics to show that when the uncharged conductor 2 (of any shape) is brought near to the charged conductor 1 (of any shape), then the electrostatic energy of the system decreases.

But this energy can be written as $U = \frac{1}{2} Q_1V_1 + \frac{1}{2} Q_2V_2 = \frac{1}{2} Q_1V_1$, since $Q_2 = 0$.

$Q_1$ remains constant while 2 is brought near. Thus, $V_1$ must decrease since $U$ decreases.

 

[sergiokapone]

Exellent! I think this is enough rigorous. Now what one can say about potential of uncharged body? From potential matrix we can conclude only $\phi_2 = V_{12} q_1$ (we know only $V_{12}>0$), but the tendense of changing $V_{12}$ remain unknown. From the special case of distant ($\ell \gg$) spheres, one know the $V_{12}$ (and, of course $\phi_2$) increasing with decreasing $\ell$, but how one can conclude the same in general case?

 

[TSny]

Exellent! I think this is enough rigorous. Now what one can say about potential of uncharged body? From potential matrix we can conclude only $\phi_2 = V_{12} q_1$ (we know only $V_{12}>0$), but the tendense of changing $V_{12}$ remain unknown. From the special case of distant ($\ell \gg$) spheres, one know the $V_{12}$ increasing with decreasing $\ell$, but how one can conclude the same in general case?

When conductor 2 is infinitely far away from conductor 1, the potential of 2 will be zero. However, when 2 is near 1, $V_{12}$ will be nonzero and positive. So, $\phi_2$ will be positive.