Charge on two spheres related to Earthing

[songoku]

Homework Statement

A positively charged rod is brought close to metallic spheres X and Y which are in contact. Which statement is correct, if Y is earthed without removing Z? (picture shown below)
a. Both X and Y will be positively charged
b. Both X and Y will be negatively charged
c. X has no charge and Y will be negatively charged
d. X will be positively charged and Y has no charge

Relevant Equations

None

Before grounding (left picture), X will be positively charged and Y will be negatively charge.

After grounding, I think electrons from Earth will flow to sphere Y and then move to sphere X so X will be neutral and Y will be negatively charged (answer C). But the answer key is D. Why?

Thanks

 

[Orodruin]

I would say the correct answer is B, but my argument is way more involved than what would count as "introductory" physics ...

 

[Steve4Physics]

X and Y are (presumably) initially uncharged, so their total charge before grounding is zero.

Grounding results in electrons flowing into Y.

Questions:
1. What is the sign of the total charge on X and Y after grounding?
2. Which of the four choices is/are consistent with this?

EDIT. Note, if you answer the above questions, you'll see why the answer cannot be D. I agree with what @Orodruin and @haruspex have said.

 

[haruspex]

I would say the correct answer is B, but my argument is way more involved than what would count as "introductory" physics ...

Yes, it is a much harder problem than was presumably intended. It is not even obvious (to me) that the two charges would have the same sign.
To get a handle on it, I considered two spheres of some small radius r, at distances $x_0, x_1$ in the same straight line from a point charge q. The grounded spheres are connected by a wire of negligible width (or grounded independently) and have charges $q_0, q_1$.
I get that for small r the charges approximate $q_i=-q\frac{r}{x_i}$.

 

[Orodruin]

Yes, it is a much harder problem than was presumably intended. It is not even obvious (to me) that the two charges would have the same sign.
To get a handle on it, I considered two spheres of some small radius r, at distances $x_0, x_1$ in the same straight line from a point charge q. The grounded spheres are connected by a wire of negligible width (or grounded independently) and have charges $q_0, q_1$.
I get that for small r the charges approximate $q_i=-q\frac{r}{x_i}$.

My argument is qualitative. With the spheres in contact, both of them are grounded at potential zero. Due to the potential satisfying the Poisson equation in the space outside the spheres, the potential will be positive everywhere. This means the derivative in the radial direction just outside any of the spheres will be positive. With the field inside the spheres being zero, this means the surface charge on both spheres is negative everywhere and thus each sphere is negatively charged.

 

[songoku]

X and Y are (presumably) initially uncharged, so their total charge before grounding is zero.

Grounding results in electrons flowing into Y.

Questions:
1. What is the sign of the total charge on X and Y after grounding?
2. Which of the four choices is/are consistent with this?

EDIT. Note, if you answer the above questions, you'll see why the answer cannot be D. I agree with what @Orodruin and @haruspex have said.

1) Negative

2) B and C

Yes, it is a much harder problem than was presumably intended. It is not even obvious (to me) that the two charges would have the same sign.
To get a handle on it, I considered two spheres of some small radius r, at distances $x_0, x_1$ in the same straight line from a point charge q. The grounded spheres are connected by a wire of negligible width (or grounded independently) and have charges $q_0, q_1$.
I get that for small r the charges approximate $q_i=-q\frac{r}{x_i}$.

My argument is qualitative. With the spheres in contact, both of them are grounded at potential zero. Due to the potential satisfying the Poisson equation in the space outside the spheres, the potential will be positive everywhere. This means the derivative in the radial direction just outside any of the spheres will be positive. With the field inside the spheres being zero, this means the surface charge on both spheres is negative everywhere and thus each sphere is negatively charged.

I got this question from friend who in grade 9. Is there maybe a possible simpler explanation to obtain the correct answer?

Thanks

 

[haruspex]

1) Negative

2) B and C
I got this question from friend who in grade 9. Is there maybe a possible simpler explanation to obtain the correct answer?

Thanks

Since the official answer is demonstrably wrong, I see no reason to suppose there is an easier solution.

 

[Orodruin]

It would however be of interest to hear the argumentation the author intends for D...

 

[songoku]

Yes, it is a much harder problem than was presumably intended. It is not even obvious (to me) that the two charges would have the same sign.
To get a handle on it, I considered two spheres of some small radius r, at distances $x_0, x_1$ in the same straight line from a point charge q. The grounded spheres are connected by a wire of negligible width (or grounded independently) and have charges $q_0, q_1$.
I get that for small r the charges approximate $q_i=-q\frac{r}{x_i}$.

Since it would be impossible for me to understand Orodruin's advance argument, I will try to understand this method.

I tried to draw free body diagram for sphere Y but it seems not making any sense. There will be electrostatic attraction force to the right with rod Z and also contact force with sphere X which is also to the right and this will result in Y being accelerated to the right which does not make sense.

How to approach this question? Thanks

 

[haruspex]

Since it would be impossible for me to understand Orodruin's advance argument, I will try to understand this method.

I tried to draw free body diagram for sphere Y but it seems not making any sense. There will be electrostatic attraction force to the right with rod Z and also contact force with sphere X which is also to the right and this will result in Y being accelerated to the right which does not make sense.

How to approach this question? Thanks

You should assume all components are being held in place by some non-conducting and uncharged apparatus.

 

[Steve4Physics]

I tried to draw free body diagram for sphere Y but it seems not making any sense. There will be electrostatic attraction force to the right with rod Z and also contact force with sphere X which is also to the right and this will result in Y being accelerated to the right which does not make sense.

It makes perfect sense. E.g. why do you think a charged comb attracts an uncharged (or oppositely charged) piece of tissue paper? The tissue paper initially accelerates towards the comb.

But, as @haruspex indicates, X and Y would typically be fixed. In a simple laboratory demo’, X and Y might be on insulated stands and Z might be hand-held. So there would be extra forces preventing movement. (Also, the forces between the charged objects would be very small.)

Thinking about free-body diagrams and forces might be of interest but it won't help you answer the original question.

Crudely guessing a possible charge distribution (making up some random values for illustration purposes only) maybe something like this is possible:

Before earthing:
Y has net charge of -10units (mainly on its right side).
X has net charge of +10units (mainly on its left side).

After earthing:
Y has net charge of -9 units (mainly on its right side).
X has net charge of +6 units (mainly on its left side).

The charge-distributions across the surfaces are non-uniform. For example X (charge +6 units) could have +8 units on its left side and-2 units on its right side. We just can’t tell without knowing the sizes/distance and doing some difficult calculations.

(Edit - typo'.)

 

[Orodruin]

After earthing:
Y has net charge of -9 units (mainly on its right side).
X has net charge of +6 units (mainly on its left side).

This would violate both spheres being negatively charged and is therefore not a possible configuration.

 

[Steve4Physics]

This would violate both spheres being negatively charged and is therefore not a possible configuration.

Yes, apologies - I forgot your (very nice) Post #5 argument.

(So option b) is the correct answer to the original question.)

 

[songoku]

You should assume all components are being held in place by some non-conducting and uncharged apparatus.

The charge-distributions across the surfaces are non-uniform. For example X (charge +6 units) could have +8 units on its left side and-2 units on its right side. We just can’t tell without knowing the sizes/distance and doing some difficult calculations

Is the calculation still possible in the scope of high school physics? If not, what should I read to be able to solve this?

Thanks

 

[haruspex]

Is the calculation still possible in the scope of high school physics? If not, what should I read to be able to solve this?

Thanks

The calculation of a charge distribution is pretty hard in most situations. E.g. for a grounded sphere near a point charge see section 3 of https://www.wtamu.edu/~cbaird/Lecture3.pdf.

But you don't need to find the actual distribution here. If I understand @Orodruin's argument in post #5, there is a general principle:
Given only a grounded body in the presence of a positive charge, the potential everywhere outside the body is positive.
@Orodruin deduces that from Poisson's equation, but I'll leave him to explain how.

If the surface charge on the body were anywhere positive, the overall potential just outside the body at that point would be a bit less than right at that point. Since the body is grounded, that would mean the potential just outside is negative, violating the principle. It follows that the surface charge is everywhere negative.

 

[Orodruin]

@Orodruin deduces that from Poisson's equation, but I'll leave him to explain how

Well, it certainly is not a high school level argument, but apart from at the rod, where the potential is positive, there is no source so what needs to be solved is the Laplace equation. A property of the Laplace equation is that it can have no local minima and since the potential tends to zero at the spheres and at infinity, having negative potential anywhere would necessarily require a local minimum to exist. Since no local minimum exists, the potential must be positive everywhere.

 

[songoku]

Thank you very much for all the help and explanation Orodruin, Steve4Physics, haruspex

 

[Orodruin]

Just the missing piece for anyone interested: Showing that no internal point can be a local minimum of the potential.

Consider the mean value $\bar f(\rho)$ of a function $f$ over the surface of a ball $B_p(\rho)$ with radius $\rho$ centered around a point $p$:
$$
\bar f(\rho) = \frac{1}{A \rho^{n-1}} \oint_{\partial B_p(\rho)} f \, dS
$$
where $A$ is the area of the unit sphere $S$ in $\mathbb R^n$. This may be rewritten as
$$
\bar f(\rho) = \frac 1A \oint_S f(\rho, \Omega) d\Omega,
$$
where we have written $f$ in polar coordinates based on $p$ and $\Omega$ are the angular coordinates parametrizing the unit sphere. Differentiating $\bar f$ leads to
$$
\bar f'(\rho) = \frac 1A \oint_S f_\rho(\rho, \Omega) d\Omega = \frac 1{A\rho^{n-1}} \oint_{\partial B_p(\rho)} \nabla f \cdot d\vec S.
$$
Applying the divergence theorem
$$
\bar f'(\rho) = \frac 1{A\rho^{n-1}} \int_{B_p(\rho)} \nabla^2 f \, dV = 0
$$
assuming that $f$ satisfies Laplace's equation. It follows that $\bar f(\rho)$ is independent of $\rho$ and in particular that $\bar f(0) = f(p)$. Thus, $f(p)$ is the mean value of $f$ over the surface of the ball, which means it cannot be smaller than (or bigger than) all of the values on the ball and therefore not be a local min (or max).