Electrostatics and current problem with 3 metallic balls

[tellmesomething]

Homework Statement

Three uncharged metallic balls of radii a, b and a respectively are connected to terminals A, B and C with the help of long thin conductors as shown in the circuit. Find changes established on each of the balls, when a steady state is reached after the switch is closed. Consider the balls to be at great distances from each other as well from the circuit and neglect internal resistance of the battery.

Relevant Equations

None

Theres a picture of the circuit attached.

I did not understand how to start the problem since we dont know the distance between the terminal A and the sphere to write the absolute potential at terminal A. So I looked at the solution and im afraid I dont understand it at all.

The teacher took the potential at terminal A to just be $\frac{kq1}{a}$

I dont understand why this is the case firstly, the potential at terminal A should be less than this potential which we get inside and at the radius of the sphere.s
Secondly the absolute potential at A should be $Potential due to sphere + V_{0}$

Then he wrote the equation
$\frac{kq1}{a} - \frac{kq2}{b} = \frac{V_{0}}{2}$

How is this correct? Shouldnt we consider the additional $V_{0}$ in the LHS and also the distances in rhe potnetial expression

 

[PhDeezNutz]

I dont understand why this is the case firstly, the potential at terminal A should be less than this potential which we get inside and at the radius of the sphere.s

Metallic balls are equipotentials.

 

[kuruman]

• The spheres are equipotentials at the potential of points A, B and C.
• Since the resistors have the same value, the potential differences are $V_{AB}=V_{BC}=\frac{1}{2}V_0.$
• The distance between spheres is said to be "great"; this means that they can be considered isolated so that their potential relative to infinity is $V_i=\dfrac{kq_i}{R_i}.$

 

[tellmesomething]

Metallic balls are equipotentials.

Oh wow, can you refer me to a link which discusses so? As far as I know field outside a metallic sphere decreases with distance from the centre no? Like it behaves like a conducting sphere. ..so to go from a point outside to another point outside work is done which would imply potential at those two points are not the same

 

[tellmesomething]

• The spheres are equipotentials at the potential of points A, B and C.
  
  
  

Does it not behave like a conducting sphere?

 

[kuruman]

Does it not behave like a conducting sphere?

It does. Why do you think it doesn't?

 

[tellmesomething]

It does. Why do you think it doesn't?

But then fields outside of a conducting sphere are not uniform. How can we say potential at all points outside is the same?

 

[kuruman]

When we say that conductors are equipotentials, we mean that every point on the conductor is at the same potential as every other point on the conductor. Of course, points outside the conductor are at different potentials. That has nothing to do with the argument.

The argument is that point A on the circuit is at the same potential as the entire sphere connected to it. The same is true for the spheres connected to points B and C.

 

[tellmesomething]

When we say that conductors are equipotentials, we mean that every point on the conductor is at the same potential as every other point on the conductor. Of course, point outside the conductor are at different potentials. That has nothing to do with the argument.

The argument is that point A on the circuit is at the same potential as the entire sphere connected to it. The same is true for the spheres connected to points B and C.

I see I was thinking of terminal A to be an outside point for the sphere. Though now it looks more complicated how can we say that the potential due to the sphere as well as the wire at terminal A is kq/a... Any hints?

 

[kuruman]

I see I was thinking of terminal A to be an outside point for the sphere. Though now it looks more complicated how can we say that the potential due to the sphere as well as the wire at terminal A is kq/a... Any hints?

Imagine switches on the spheres connecting the three spheres to points A, B and C. We are in the steady state and the pshres are fully charged. Now open all three switches.

Question: What happens?
Answer: The charges are trapped on the spheres. The potential differences between adjacent spheres are $$V_{AB}=\frac{kq_B}{R_B}-\frac{kq_A}{R_A}~;~~V_{BC}=\frac{kq_C}{R_C}-\frac{kq_B}{R_B}.$$The circuit is isolated from the spheres and $V_{AB}=V_{BC}=\frac{1}{2}V_0.$

Now close the switches.
Questions:
Do the charges on the spheres change? If so how and why?
Do the potential differences between spheres change? If so how and why?
Do the potential differences between points A, B and C change? If so how and why?

 

[tellmesomething]

Imagine switches on the spheres connecting the three spheres to points A, B and C. We are in the steady state and the pshres are fully charged. Now open all three switches.

Question: What happens?
Answer: The charges are trapped on the spheres. The potential differences between adjacent spheres are $$V_{AB}=\frac{kq_B}{R_B}-\frac{kq_A}{R_A}~;~~V_{BC}=\frac{kq_C}{R_C}-\frac{kq_B}{R_B}.$$The circuit is isolated from the spheres and $V_{AB}=V_{BC}=\frac{1}{2}V_0.$

I get what you say but the potential difference between the two spheres, you are implying that potentials at terminal A is $\frac{kq1}{a}$ abd potnetial at terminal B is $\frac{kq2}{b}$

I dont get how you get the above, potential at a point p1 and p2 inside is indeed $\frac{kq1}{a}$ $\frac{kq2}{b}$

But at a point outside of a sphere connected by a conducting wire...how do you figure that out? The charge must obviously distribute inside the conducting wire as well. .and the potential at terminal A must be due to both the sphere as well as the wire connecting the sphere to the terminal A

 

[kuruman]

I get what you say but the potential difference between the two spheres, you are implying that potentials at terminal A is $\frac{kq1}{a}$ abd potnetial at terminal B is $\frac{kq2}{b}$

No. Electrostatic potential is not the same as electrostatic potential difference. The potential the charged spheres is $\frac{kq}{r}$ relative to infinity where the zero of potential is defined. In other words, $\frac{kq}{r}$ is the electrostatic potential difference between a sphere an infinity. Here you know that sphere A is at the same potential as point A but you don't know the electrostatic potential at point A relative to infinity. That is why you have to consider differences of potential.

But at a point outside of a sphere connected by a conducting wire...how do you figure that out? The charge must obviously distribute inside the conducting wire as well. .and the potential at terminal A must be due to both the sphere as well as the wire connecting the sphere to the terminal A

The charge accumulating in the wires is negligible because they have negligible capacitance. That is not true for the spheres. You can view them as the inner sphere of a spherical capacitor with the outer shell having infinite radius (see equivalent diagram below.)

 

[tellmesomething]

No. Electrostatic potential is not the same as electrostatic potential difference. The potential the charged spheres is $\frac{kq}{r}$ relative to infinity where the zero of potential is defined. In other words, $\frac{kq}{r}$ is the electrostatic potential difference between a sphere an infinity. Here you know that sphere A is at the same potential as point A but you don't know the electrostatic potential at point A relative to infinity. That is why you have to consider differences of potential.



View attachment 349460

If i know that the absolute potential of the sphere is kq/r and I know that the potential of the sphere and point A is the same then wouldnt that mean the potential at point A is also kq/r?
Though I guess thats what I dont understand how did you conclude that potential at the sphere is same as the potential at point A

 

[kuruman]

Though I guess thats what I dont understand how did you conclude that potential at the sphere is same as the potential at point A

One more time, each sphere and all the connecting wires attached to them are equipotential. In the figure on the right all the conducting material within each colored shape is at the same potential. The vacuum is, of course, excluded.

• The potential difference between all the conductors within red and all the conductors within yellow is $\frac{1}{2}V_0.$
• The potential difference between all the conductors within yellow and all the conductors within green is $\frac{1}{2}V_0.$
• The potential difference between all the conductors within red and all the conductors within green is $V_0.$

 

[tellmesomething]

View attachment 349493One more time, each sphere and all the connecting wires attached to them are equipotential. In the figure on the right all the conducting material within each colored shape is at the same potential. The vacuum is, of course, excluded.

• The potential difference between all the conductors within red and all the conductors within yellow is $\frac{1}{2}V_0.$
• The potential difference between all the conductors within yellow and all the conductors within green is $\frac{1}{2}V_0.$
• The potential difference between all the conductors within red and all the conductors within green is $V_0.$

Okay I think I get it now..
So to answer this

Now close the switches.
Questions:
Do the charges on the spheres change? If so how and why?
Do the potential differences between spheres change? If so how and why?
Do the potential differences between points A, B and C change? If so how and why?

NoI it will not change..?

The potential difference will not change since no charge from the sphere flows out..

No the potential difference between A B and C does not change as well..

 

[Steve4Physics]

Though I guess thats what I dont understand how did you conclude that potential at the sphere is same as the potential at point A

The question tells you that the system has reached a 'steady state'. That means each sphere has a constant charge. So each sphere's connecting wire has zero current.

If the potentials at point A and its sphere were different, a current would flow between them. Since the current is zero, the potential difference between point A and its sphere is zero. Therefore point A and its sphere are at the same potential - as nicely shown in red in @kuruman's Post #14 diagram.

 

[tellmesomething]

The question tells you that the system has reached a 'steady state'. That means each sphere has a constant charge. So each sphere's connecting wire has zero current.

I see so if it wasnt mentioned that it has reached the steady state, we could not have concluded this?

 

[tellmesomething]

Also i get two equations following the above discussion $\frac{kq1}{a}-\frac{kq2}{b}=\frac{V_0}{2}$
$\frac{kq2}{b}-\frac{kq3}{a}=\frac{V_0}{2}$ can someone hint at a third equation i can derive....? to successully solve this ques

 

[Steve4Physics]

I see so if it wasnt mentioned that it has reached the steady state, we could not have concluded this?

Correct

But note, for some questions, the term 'steady state' may not be stated explicitly. It might be implied from the context and/or wording.

 

[kuruman]

can someone hint at a third equation i can derive....? to successully solve this ques

What is the total charge in all the conductors before the battery is connected?
What is the total charge in all the conductors in the steady state?

 

[Steve4Physics]

Also i get two equations following the above discussion $\frac{kq1}{a}-\frac{kq2}{b}=\frac{V_0}{2}$
$\frac{kq2}{b}-\frac{kq3}{a}=\frac{V_0}{2}$ can someone hint at a third equation i can derive....? to successully solve this ques

Apart from the polarity (+/-) of the cell, the circuit is symmetrical because the outer spheres have equal radii and the 2 resistors are equal. What can you deduce about $q_1$ and $q_3$?

 

[tellmesomething]

What is the total charge in all the conductors before the battery is connected?
What is the total charge in all the conductors in the steady state?

The total charge before the battery is connected is 0. After connecting battery it must be 0 as well ofcourse. Though sorry if this is a very elementary question but charge is not flowing from the circuit ? so if the sphere was initially neutral how would it even have a charge? Sure the charges might redistribute but net charge would remain 0..

 

[kuruman]

Apart from the polarity (+/-) of the cell, the circuit is symmetrical because the outer spheres have equal radii and the 2 resistors are equal. What can you deduce about $q_1$ and $q_3$?

The relation between $q_1$ and $q_3$ can also be obtained by subtracting the two equations.

Sure the charges might redistribute but net charge would remain 0..

Yes.

 

[tellmesomething]

Apart from the polarity (+/-) of the cell, the circuit is symmetrical because the outer spheres have equal radii and the 2 resistors are equal. What can you deduce about $q_1$ and $q_3$?

I'll think about this approach..

 

[tellmesomething]

The relation between $q_1$ and $q_3$ can also be obtained by subtracting the two equations.

Yes.

All the spheres are uncharged initially and then while solving we are supposed to assume that they are charged? But how is this valid if no charge from the Sphere A flows to the circuit then the net charge on sphere A is 0. Similarly net charge on sphere B and sphere C is 0 even when the battery is connected..How are we having the discussion on electric potentials? How is this charge q1,q2 and q3 developing when there is no transfer of charge from any of the spheres? What i mean is net charge on each sphere individually comes out to be 0..

 

[kuruman]

Charge flows in the transient state, immediately after the battery is connected to the time when steady state is reached. It's like charging a capacitor except that here you have three of them.

 

[tellmesomething]

Charge flows in the transient state, immediately after the battery is connected to the time when steady state is reached. It's like charging a capacitor except that here you have three of them.

I see that makes sense. Many thanks for this whole discussion @kuruman and @Steve4Physics

 

[Steve4Physics]

The relation between $q_1$ and $q_3$ can also be obtained by subtracting the two equations.

FWIW...

$\frac{kq_1}{a}-\frac{kq_2}{b}=\frac{V_0}{2} ~~(1)$
$\frac{kq_2}{b}-\frac{kq_3}{a}=\frac{V_0}{2} ~~(2)$

Subtracting (say (2) from (1)) gives:
$\frac{kq_1}{a}- 2\frac{kq_2}{b} +\frac{kq_3}{a} = 0$
leaving an equation containing all three charges:
$\frac{q_1}{a}- 2\frac{q_2}{b} +\frac{q_3}{a} = 0$

Adding (1) and (2) gives:
$\frac{kq_1}{a}-\frac{kq_3}{a}=V_0$
$q_1 - q_3 = \frac {aV_0}k$
Although this gives a relationship betwee $q_1$ and $q_3$ it's not the key relationship that the (anti)symmetry imposes.

 

[kuruman]

Yes, I should stop trying to solve problems in my head before having my morning coffee.

Perhaps a more instructive formulation of this problem might be to ground the right side of the circuit as shown on the right. The spherical capacitance with the outer sphere at infinity is $C_i=4\pi\epsilon_0~R_i.$ Then

1. The electric potential at C is the same as at infinity, $V_C=0.$ So the charge on the sphere at C is $$Q_C=C_C~V_C=4\pi\epsilon_0~R_C\times 0=0.$$
2. The electric potential at B is $~V_B=\frac{1}{2}V_0~$ relative to infinity. The charge on the sphere at B is $$Q_B=C_B~V_B=C_B=4\pi\epsilon_0~R_B\times\frac{1}{2}V_0=2\pi\epsilon_0~R_BV_0.$$
3. The electric potential at A is $~V_A=V_0~$ relative to infinity. The charge on the sphere at A is $$Q_A=C_A~V_A=4\pi\epsilon_0~R_AV_0.$$

We can treat the original problem the same way. If we assume that the potential at C relative to infinity is $V_C$. then
$Q_C=4\pi\epsilon_0~a V_C$
$Q_B=4\pi\epsilon_0~b( V_C+\frac{1}{2}V_0)$
$Q_A=4\pi\epsilon_0~a (V_C+V_0)$
Charge conservation requires that
$$0=Q_A+Q_B+Q_C=4\pi\epsilon_0\left[a V_C+b\left( V_C+\frac{1}{2}V_0\right)+a (V_C+V_0)\right]\implies V_C=-\frac{1}{2}V_0.$$Substitute in the three equations to get,$$Q_C=-2\pi\epsilon_0~a V_0~;~~Q_B=0~;~~Q_A=2\pi\epsilon_0~a V_0.$$

 

[tellmesomething]

Yes, I should stop trying to solve problems in my head before having my morning coffee.

View attachment 349521Perhaps a more instructive formulation of this problem might be to ground the right side of the circuit as shown on the right. The spherical capacitance with the outer sphere at infinity is $C_i=4\pi\epsilon_0~R_i.$ Then

1. The electric potential at C is the same as at infinity, $V_C=0.$ So the charge on the sphere at C is $$Q_C=C_C~V_C=4\pi\epsilon_0~R_C\times 0=0.$$
2. The electric potential at B is $~V_B=\frac{1}{2}V_0~$ relative to infinity. The charge on the sphere at B is $$Q_B=C_B~V_B=C_B=4\pi\epsilon_0~R_B\times\frac{1}{2}V_0=2\pi\epsilon_0~R_BV_0.$$
3. The electric potential at A is $~V_A=V_0~$ relative to infinity. The charge on the sphere at A is $$Q_A=C_A~V_A=4\pi\epsilon_0~R_AV_0.$$

We can treat the original problem the same way. If we assume that the potential at C relative to infinity is $V_C$. then
$Q_C=4\pi\epsilon_0~a V_C$
$Q_B=4\pi\epsilon_0~b( V_C+\frac{1}{2}V_0)$
$Q_A=4\pi\epsilon_0~a (V_C+V_0)$
Charge conservation requires that
$$0=Q_A+Q_B+Q_C=4\pi\epsilon_0\left[a V_C+b\left( V_C+\frac{1}{2}V_0\right)+a (V_C+V_0)\right]\implies V_C=-\frac{1}{2}V_0.$$Substitute in the three equations to get,$$Q_C=-2\pi\epsilon_0~a V_0~;~~Q_B=0~;~~Q_A=2\pi\epsilon_0~a V_0.$$

I could not follow the capacitance approach since we haven't yet covered that. Though its supposed to be our next unit. I really appreciate this and will come back to it after ive gained enough context about capacitance . thankyou