How does metal sphere determine the potential of the plane by symmetry?

[yucheng]

Homework Statement:

See https://physics.stackexchange.com/questions/275950/uniqueness-theorem-in-uniform-electric-field

The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. Then $V$ does not go to zero but rather far from the sphere the field is $E_0 \hat{z}$ we thus have $$v \to -E_0z + C.$$

Relevant Equations:

N/A

The answer given states that:

The entire x-y plane is obviously at the same potential since all the fields are strictly perpendicular to it (draw a diagram if youre confused). Since we choose the sphere to be at potential zero, the point on the sphere which cuts the x-y plane is also at zero potential, and hence the entire plane is at zero by the definition chosen

My concerns are:
1. Which x-y plane? $z=0$ only, or $-R \leq z \leq R$?
2. Won't the argument thus apply to the whole 'slab' with thickness $2R$ (along z-axis, diameter of sphere), and therefore along the whole slab, it would be of zero potential?

Essentially, why will Griffiths claim that 'The sphere is an equipotential we can set it to zero. Then by symmtery the entire $xy$ plane is at potential zero. ' ?

Thanks in advance!

 

[kuruman]

By symmetry, the entire $xy$ plane at $z = 0$ will be an equipotential at V = 0. Also, the entire volume occupied by the sphere will be an equipotential at V = 0 because the sphere is a conductor. I don't know what you mean by "slab".

 

[Keith_McClary]

It is $z=0$ because that is the same under the symmetry of reversing the polarity.

 

[yucheng]

symmetry of reversing the polarity.

May I know what this means?

 

[yucheng]

By symmetry, the entire $xy$ plane at $z = 0$ will be an equipotential at V = 0. Also, the entire volume occupied by the sphere will be an equipotential at V = 0 because the sphere is a conductor. I don't know what you mean by "slab".

By which symmetry? How does this make the $xy$ plane at $z=0$ be equipotential at V=0?

By slab, I meant volume bounded by the $xy$ plane at $z=R$ and $z=-R$.

 

[Steve4Physics]

I’ll chip in.

The xy plane is the sphere's equatorial plane.

The positive charge distribution in the sphere’s northern hemisphere is a reflection of the negative charge distribution in the southern hemisphere. That’s the key symmetry.

As a result, outside the sphere the field lines are perpendicular to the xy (equatorial) plane. So we have this: https://i.stack.imgur.com/MdsGv.png (source: https://physics.stackexchange.com/questions/72390/understand-equations-of-a-conducting-sphere)

Just outside the equator, the field lines are tangential to the sphere - so no work is done moving from a point on the equator radially outwards. That means the xy plane has the same potential as the sphere (which has arbitrarily been chosen to be zero).

If that helps.

But I have no idea what “$v \to -E_0z + C$” means!

 

[Delta2]

But I have no idea what “v→−E0z+C” means

If I understood well what is being said at the link at stackexchange, it wants to say that the potential far way from the sphere (for distances r>>R , R the radius of the sphere), the potential is approximated by that expression.

 

[Delta2]

I guess far away from the sphere, the influence of the sphere's charge distribution is neglected (the sphere behaves like a point charge with total charge +q-q=0, where q the charge at each hemisphere of the sphere) and we are left only with the external electric field $E_0$.

 

[Steve4Physics]

I guess far away from the sphere, the influence of the sphere's charge distribution is neglected and we are left only with the external electric field $E_0$.

Aha. Thanks @Delta2. I was confused by the use of lower case "$v$" which is just a typo' and should be upper case "$V$". I must be in need of a coffee.

 

[Delta2]

@yucheng If you want an answer that uses more math, the potential at a point $(x_0,y_0,0)$ at the z=0 plane will be
$$V(x_0,y_0,0)=\iiint_K\frac{\rho(x,y,z)}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-0)^2}}dxdydz$$ where K is the volume of the sphere and $\rho$ the volume charge density). Now because the induced charge density is such that is antisymmetric with z=0, that is $\rho(x,y,z)=-\rho(x,y,-z)$. You can break this integral into the sum of two integrals one for z>0 and one for z<0 and due to the antisymmetric property (and because the denominator contains $z^2$ which is symmetric for z=0, that is $z^2=(-z)^2$), their sum will be zero.

 

[kuruman]

@yucheng If you want an answer that uses more math, the potential at a point $(x_0,y_0,0)$ at the z=0 plane will be
$$V(x_0,y_0,0)=\iiint_K\frac{\rho(x,y,z)}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-0)^2}}dxdydz$$ where K is the volume of the sphere and $\rho$ the volume charge density). Now because the induced charge density is such that is antisymmetric with z=0, that is $\rho(x,y,z)=-\rho(x,y,-z)$. You can break this integral into the sum of two integrals one for z>0 and one for z<0 and due to the antisymmetric property (and because the denominator contains $z^2$ which is symmetric for z=0, that is $z^2=(-z)^2$), their sum will be zero.

There is no volume charge density. It is a metal sphere.

 

[Delta2]

There is no volume charge density. It is a metal sphere.

In my opinion there is but it is expressed as a product of the dirac delta function (everywhere zero except at the points r=R where it becomes infinite) and the surface charge density (which I assume we all agree that is well defined).

 

[Delta2]

In any case we can define a surface integral with surface charge density (instead of a volume integral with volume charge density) and using similar logic as post #10 we can still conclude that this surface integral will be zero for the potential for points $(x_0,y_0,0)$.

 

[kuruman]

By which symmetry? How does this make the $xy$ plane at $z=0$ be equipotential at V=0?

By slab, I meant volume bounded by the $xy$ plane at $z=R$ and $z=-R$.

Consider person A looking at the sphere so that the E-field is pointing to his right and person B looking from the opposite side with the E-field pointing to his left. They are looking at the same physical situation but from different directions. All magnitudes are the same but what's right (positive) for one person is left (negative) for the other. Mathematically, $V_\text{left}=-V_\text{right}$ or more simply $V(-z)=-V(z)$. Now the z-component of the electric field (cylindrical coordinates) in the region $z>0$ is $$E_z(r,z)=-\frac{\partial V(z)}{\partial z}$$ In the region $z<0$ it is $$E_z(r,-z)=-\frac{\partial V(-z)}{\partial (-z)}=-\frac{\partial (-V(z))}{\partial (-z)}=-\frac{\partial V(z)}{\partial z}=E_z(r,z)$$This says that the z-component of the electric field is continuous across the $z=0$ plane. Since the potential on one side of the plane is the negative of the potential on the other and its derivative is non-zero, it follows that $V(r,0)=0.$

It also says that the equipotential slab at zero potential that you pictured is absurd. The z-component of the electric field is continuous across any plane parallel to the $z=0$ plane. If your slab picture were correct, then you would have $E(r, z=R)=E(r, z=-R)=0$ for $r>R$ where there is vacuum. This means that there's a discontinuity in the electric field without a surface charge density. In other words, electric field lines would start and stop in the middle of a vacuum which violates Gauss's law.

 

[kuruman]

In my opinion there is but it is expressed as a product of the dirac delta function (everywhere zero except at the points r=R where it becomes infinite) and the surface charge density (which I assume we all agree that is well defined).

Yes, of course, your opinion is correct. When I saw $\rho(x,y,z)$ instead of $\sigma(\theta,\phi)$ or $\delta(r-R)$, I thought that you may have misread the statement of the problem in the stackexchange link.

 

[yucheng]

It also says that the equipotential slab at zero potential that you pictured is absurd. The z-component of the electric field is continuous across any plane parallel to the $z=0$ plane. If your slab picture were correct, then you would have $E(r, z=R)=E(r, z=-R)=0$ for $r>R$ where there is vacuum. This means that there's a discontinuity in the electric field without a surface charge density. In other words, electric field lines would start and stop in the middle of a vacuum which violates Gauss's law.

Looking back at what I've done 6 months ago, I think my mistake was that I failed to reason by first principles... Your explanation given was quite nice actually!

I do think that the most important property of the potential is superposition.

Let me just summarize:
We first use the symmetry of the sphere (how the charges in one hemisphere is equal but opposite to the other) to conclude that there exists an equipotential plane, such that the potential function of the induced charges on the sphere, is zero; specifically, the equatorial plane.

The constancy of the electric field means that the potential of the imposed field is linear with respect to the z-axis which is perpendicular to the equatorial plane; we can translate the potential of the field (by adding an arbitrary constant) such that it coincides with that of the equatorial plane, i.e. 0. Since any plane perpendicular to the z-axis is an equipotential surface, the potential of the electric field is constant on the equatorial plane hence takes on 0 on the whole plane.

Then we superimpose both fields, the equatorial plane is still 0.

Finally we solve the boundary value problem.

Thanks all!