# Field confusion in conductors

• B
• rudransh verma

#### rudransh verma

Gold Member
If there is a spherical conductor like this with excess charges on its surface then this is the field it sets up. Each e- would have there own radial field of lines. And all the e-s would exert a force on each charge and the net field on each charge and thus net force would be zero. All the e-s will be at rest. I tried to show this net effect of all field vectors on one top centre charge.

If I am correct then i want to ask what does it mean when we say the electric fields (vectors) are perpendicular to the surface of a charged conductor. Which field is this? Is this the net field vector at each point on the surface due to all the charges. Or is this field due to single e- that is perpendicular.

#### Attachments

• AC75E937-ACC3-4E42-A13F-A0E1B03523C1.jpeg
54.3 KB · Views: 57
Last edited:

The net field is perpendicular to the surface and your drawing hints it is the case.

The net field is perpendicular to the surface and your drawing hints it is the case.
I have drawn fields due to all charges on one charge. The resultant of all is one field perpendicular to the surface. And in this way all resultant fields are perpendicular on the entire surface. Charges don’t move further due to this resultant because they are on boundary now.
We say net field on each charge is zero ?

There is only one field. In this case it's very easy to calculate. Due to symmetry the electrostatic potential is radially symmetric, i.e., it only depends on ##r=|\vec{x}|## (having the center of the sphere with radius ##a## at the origin of our coordinate system). Using spherical coordinates it fulfills the Laplace equation everywhere except on the spherical shell, where the charge sits. This means
$$\Delta \Phi=0.$$
Since ##\Phi=\Phi(r)## this means
$$\frac{1}{r}[r \Phi(r)]''=0.$$
Multiplying with ##r## and integrating twice wrt. ##r## you get
$$r \Phi(r)=C_1 r+C_2 \; \Rightarrow \; \Phi(r)=C_1+\frac{C_2}{r}.$$
Since there's nothing at the origin which could make the potential singular we must have ##C_2=0## in the region ##r<a##, i.e.,
$$\Phi(r)=C_1 \quad \text{for} \quad r<a.$$
Then we want the potential to go to 0 at infinity, which chooses the arbitrary overall additive constant. So we have
$$\Phi(r)=\frac{C_2}{r} \quad \text{for} \quad r>a.$$
Further ##\Phi## should be continuous at ##r=a##, such that ##C_1=C_2/a##.

The electric field thus is
$$\vec{E}=-\vec{\nabla} \Phi=\begin{cases} \vec{0} & \text{for} \quad r<a \\ C_2 \vec{r}/r^3 &\text{for} \quad r>a. \end{cases}$$
Now take an arbitrary spherical shell ##K_R## with radius ##R>a## with the center in the origin. According to Gauss's Law in integral form you have
$$\frac{Q}{\epsilon_0}=\int_{S_R} \mathrm{d}^2 \vec{x} \cdot \vec{E}.$$
In standard spherical coordinates we have
$$\mathrm{d}^2 \vec{x} = \mathrm{d} \vartheta \mathrm{d} \varphi R^2 \sin \vartheta.$$
Thus the surface integral is
$$\frac{Q}{\epsilon_0}=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi R^2 \sin \vartheta \frac{C_2}{R^2} = 4 \pi \; \Rightarrow \; C_2=\frac{Q}{4 \pi \epsilon_0}.$$
To check that this indeed describes a homogeneously charged spherical shell we calculate the jump of the normal component of ##\vec{E}## across the surface. The normal component is of course
$$E_r=\begin{cases} 0 &\text{for} \quad r<a \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for} \quad r>a. \end{cases}$$
The said jump thus is
$$\frac{\sigma}{\epsilon_0}=E_r(a+0^+)-E_r(a-0^+)=\frac{Q}{4 \pi \epsilon_0 a^2}=\text{const}.$$
So this solves your problem completely.

There is only one field.
I meant there are resultant field vectors all around the conductor due to all charges. One field around the conductor. These resultant field vectors all push the e-s at the boundary.

pardon me but I don’t understand the maths. I am a undergraduate.
@vanhees71 @Gordianus
Charges don’t move further due to these resultant vectors because they are on boundary now.
We say net field on each charge is zero.
What do you guys think ?

What's the level of math, I'm allowed to use? The thread is labeled I!

What's the level of math, I'm allowed to use? The thread is labeled I!
Just use high schoool maths.
The big dark arrows are the resultant field vectors. This resultant field cannot push the e-s out. We say net field on each e- is zero.

#### Attachments

• image.jpg
50.5 KB · Views: 47
Last edited:
I'm not sure whether your drawing is correct. Inside the sphere there shouldn't be any field lines, because there the electric field is 0.

Using what I learned in high school, I'd have to do the integral solution given the homogeneous surface-charge distribution on the sphere, i.e., letting ##\vec{x}=r \vec{e}_z## (because of symmetry you know that ##\Phi(\vec{z})## only depends on ##r=|\vec{x}|## and I can choose my coordinate system as conveniently as possible, and when using spherical coordinates it's most convenient to choose the only preferred direction in the problem, which is ##\vec{x}## in direction of the ##z## axis, i.e., the polar axis of the spherical coordinate system)
$$\Phi(\vec{x}=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi a^2 \sin \vartheta \frac{\sigma}{4 \pi \epsilon_0 \sqrt{r^2+a^2-2 r a \cos \vartheta}}.$$
The ##\varphi## integral delivers a factor ##2 \pi##. The ##\vartheta## integral can be done via substitution of ##u=\cos \vartheta##, ##\mathrm{d} u=-\sin \vartheta \mathrm{d} \vartheta##:
$$\Phi(\vec{x})=\frac{\sigma a^2}{2\epsilon_0} \int_{-1}^{1} \mathrm{d} u \frac{1}{\sqrt{r^2+a^2-2 r a u}}.$$
The integral is easily done
$$\Phi(\vec{x})=-\frac{\sigma a}{2 r \epsilon_0} \left [\sqrt{r^2+a^2-2 r a u} \right]_{u=-1}^{u+1}.$$
Now you have to distinguish the cases ##r<a## and ##r>a##.

For ##r<a## you get
$$\Phi(\vec{x})=-\frac{\sigma a}{2 \epsilon_0 r} [(a-r)-(r+a)]=\frac{\sigma a}{\epsilon_0}=\frac{Q}{4 \pi \epsilon_0 a}.$$
For ##r>a## you get
$$\Phi(\vec{x}0=-\frac{\sigma a}{2 \epsilon_0 r} [(r-a)-(r+a)]=\frac{\sigma a^2}{\epsilon_0 r}=\frac{Q}{4 \pi \epsilon_0 r},$$
which is of course the same solution as given above via the differential equation.

If there is a spherical conductor like this with excess charges on its surface then this is the field it sets up. Each e- would have there own radial field of lines. And all the e-s would exert a force on each charge and the net field on each charge and thus net force would be zero. All the e-s will be at rest. I tried to show this net effect of all field vectors on one top centre charge.

If I am correct then i want to ask what does it mean when we say the electric fields (vectors) are perpendicular to the surface of a charged conductor. Which field is this? Is this the net field vector at each point on the surface due to all the charges. Or is this field due to single e- that is perpendicular.
The E field just outside the surface is always perpendicular to a tangent (line or plane) at that point on the surface. It is the net E field due to all the charges.

Each charge contributes its own field, just as though it was the only charge.

This is true inside the conductor as well as outside. The net field inside the metal is zero due to the joint E fields of all the charges on the surface. In reality there is a near-infinite number of E fields inside the sphere, all adding to net-zero E field.

rudransh verma
The E field just outside the surface is always perpendicular to a tangent (line or plane) at that point on the surface. It is the net E field due to all the charges.

Each charge contributes its own field, just as though it was the only charge.

This is true inside the conductor as well as outside. The net field inside the metal is zero due to the joint E fields of all the charges on the surface. In reality there is a near-infinite number of E fields inside the sphere, all adding to net-zero E field.
You say There are infinite number of fields inside the sphere due to all surface charges. But net field inside is zero. How is net field zero. Which field cancels this internal field due to all surface charges?
Is it the field generated due to separation of positive atoms and e-s inside which cancels the field due to outside surface charges.

#### Attachments

• E84A80EC-5835-407C-9FFD-66650B7DD89D.jpeg
51 KB · Views: 42
Last edited:
You say There are infinite number of fields inside the sphere due to all surface charges. But net field inside is zero. How is net field zero. Which field cancels this internal field due to all surface charges?
Is it the field generated due to separation of positive atoms and e-s inside which cancels the field due to outside surface charges.
If you have zero net charge on your sphere's surface then the conduction electrons are randomly dispersed inside the sphere. There is no E field anywhere inside & outside the sphere.

If you have a net charge on the sphere then all that net charge is evenly dispersed about the surface.

Each charge q radiates its own E field = kq/r^2 omnidirectionally. The net effect of all these charges is to produce a net zero E field inside the sphere. Outside the sphere Gauss' law gives you the E field at any distance away from the surface.

This is all exactly analogous to the g field inside a uniformly massive spherical shell of mass M. In this case every iota of mass in the shell contributes to gravitational pull on a test mass inside the shell but together they cancel each other out. Outside the shell the g field = GM/r^2.

Gravity force and the electric field go as 1/r^2. Exactly analogous.

Each charge q radiates its own E field = kq/r^2 omnidirectionally. The net effect of all these charges is to produce a net zero E field inside the sphere. Outside the sphere Gauss' law gives you the E field at any distance away from the surface.
Net field inside at all points becomes zero due to all the surface charges because well the charges are all around. And the same charges produce a net field just outside the sphere which could be found out using gauss law.

@rude man something like this ? Inside the net field becomes zero at all points. I tried to show 3 points. All the charges on surface make zero internal field but outside the surface the same charges make perpendicular fields.

#### Attachments

• 404ED74A-30BB-4BCD-8FBD-9D46950C7326.jpeg
54.2 KB · Views: 44
And all the e-s would exert a force on each charge and the net field on each charge and thus net force would be zero
The resultant force of each charge can be zero, but the charge accumulated on the surface of the sphere is indeed being pushed by the radial electric field.