Electric field experienced by a uniformly charged spherical shell due to itself

[Likith D]

The electric field experienced by the points on the surface of the shell is put out as
KQ/R^2
where Q is charge on shell and R is radius of shell...

But the gaussian surface corresponding to the case intersects the sphere, which means there are non-infinitesimal charge quantity sitting on the surface, which shouldn't be so...
So, I needed some other way, to prove it if it is possible...

 

[vanhees71]

I do not understand, what you want to achieve. Of course, the electric field of a uniformly charged spherical shell is the Coulomb field of the total charge for $r>R$ ($R$: radius of the shell), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0 r^3} \vec{r} \quad \text{for} \quad r=|\vec{r}|>R,$$
while for $r<R$ there is no field at all.

The normal component of the electric field jumps, as it must be, and the jump is given by the surface-charge density (modulo the SI conversion factor):
$$\vec{n} \cdot (\vec{E}|_{r=R+\epsilon}-\vec{E}|_{r=R-\epsilon})=\frac{Q}{4 \pi \epsilon_0 R^2}.$$
Of course, there sits a finite total charge on the surface. Why shouldn't it?

 

[Likith D]

I do not understand, what you want to achieve. Of course, the electric field of a uniformly charged spherical shell is the Coulomb field of the total charge for $r>R$ ($R$: radius of the shell), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0 r^3} \vec{r} \quad \text{for} \quad r=|\vec{r}|>R,$$
while for $r<R$ there is no field at all.

The normal component of the electric field jumps, as it must be, and the jump is given by the surface-charge density (modulo the SI conversion factor):
$$\vec{n} \cdot (\vec{E}|_{r=R+\epsilon}-\vec{E}|_{r=R-\epsilon})=\frac{Q}{4 \pi \epsilon_0 R^2}.$$
Of course, there sits a finite total charge on the surface. Why shouldn't it?

Refer to : https://www.physicsforums.com/index.php?threads/99419/

 

[Jano L.]

The electric field experienced by the points on the surface of the shell is put out as
KQ/R^2
where Q is charge on shell and R is radius of shell...

But the gaussian surface corresponding to the case intersects the sphere, which means there are non-infinitesimal charge quantity sitting on the surface, which shouldn't be so...
So, I needed some other way, to prove it if it is possible...

In such case the Gauss law

electric flux through surface = charge inside the surface

cannot be used. This is because if charge is distributed on a 2D surface, the surface is a place of electric field discontinuity and neither the electric flux nor the "electric charge inside" have obviously correct definition.

The charges on the surface do experience definite electric field though, it just cannot be determined from the Gauss law. In case of uniformly charged sphere, this electric field acting on a charge that is part of the surface is half of the field just above the surface, i.e. $KQ/(2R^2)$.

This result can be found either by direct integration or (much more easily) by a relatively well-known trick where the field just above the surface (KQ/R^2) is expressed as sum of the field due to the local patch just below the point (field of charged surface $\sigma/(2\epsilon_0) = KQ/(2R^2)$ and the field due to the rest of the sphere, which has the same magnitude and direction. The charge that is in the surface experiences only the field due to the rest of the sphere and its magnitude is $KQ/(2R^2)$.

 

[Likith D]

In such case the Gauss law

electric flux through surface = charge inside the surface

cannot be used. This is because if charge is distributed on a 2D surface, the surface is a place of electric field discontinuity and neither the electric flux nor the "electric charge inside" have obviously correct definition.

The charges on the surface do experience definite electric field though, it just cannot be determined from the Gauss law. In case of uniformly charged sphere, this electric field acting on a charge that is part of the surface is half of the field just above the surface, i.e. $KQ/(2R^2)$.

This result can be found either by direct integration or (much more easily) by a relatively well-known trick where the field just above the surface (KQ/R^2) is expressed as sum of the field due to the local patch just below the point (field of charged surface $\sigma/(2\epsilon_0) = KQ/(2R^2)$ and the field due to the rest of the sphere, which has the same magnitude and direction. The charge that is in the surface experiences only the field due to the rest of the sphere and its magnitude is $KQ/(2R^2)$.

Actually, I had predicted the same... but because some textbooks state that the E field experienced by the shell on itself is actually KQ/r^2, I had to confirm my doubts...
This was very helpful, thank you!

 

[PumpkinCougar95]

This result can be found either by direct integration or (much more easily) by a relatively well-known trick where the field just above the surface (KQ/R^2) is expressed as sum of the field due to the local patch just below the point (field of charged surface σ/(2ϵ0)=KQ/(2R2)σ/(2ϵ0)=KQ/(2R2)\sigma/(2\epsilon_0) = KQ/(2R^2) and the field due to the rest of the sphere, which has the same magnitude and direction. The charge that is in the surface experiences only the field due to the rest of the sphere and its magnitude is KQ/(2R2)KQ/(2R2)KQ/(2R^2).

Another method is to think of a disk(very small) with $-\sigma$ charge density, and superimposing that disk on the sphere.

Edit: This is also useful when you have to calculate the Electric field due to the rest of the sphere at some slight distance away.

 

[Likith D]

Another method is to think of a disk(very small) with $-\sigma$ charge density, and superimposing that disk on the sphere.

Edit: This is also useful when you have to calculate the Electric field due to the rest of the sphere at some slight distance away.

Explanation of that will be very helpful...
how would one superimpose a disc on a sphere, and how would that turn out to be helpful in calculating the electric field?
sounds interesting, though not as obvious to me as you expected

 

[PumpkinCougar95]

Just like you use the superposition principle to find the net $E$ field for a system of charges. You can superimpose a small disk on a sphere.

The Electric field due to a spherical shell at a distance $\delta$ away from the shell is $$ E=\frac {Q} {4 \pi \epsilon (R+\delta)^2} $$

And for a disk Electric field at a distance $\delta$ away is $$ E = \frac {\sigma} {2\epsilon} (1 - \frac {\delta} { \sqrt{ \delta^2+r^2 }}) $$ r:radius of disk(small)

Now if you place a disk with $-\sigma$ charge density at the surface of the shell you get the same field as that due to a shell with a hole.

$$ E=\frac {Q} {4 \pi \epsilon (R+\delta)^2} - \frac {\sigma} {2\epsilon} (1 - \frac {\delta} { \sqrt{ \delta^2+r^2 }})$$ where $\sigma = \frac {Q} {4 \pi R^2}$

For $\delta = 0$ you get $E = \frac {\sigma} {2 \epsilon}$ I think this should work even when $\delta =/= 0$

 

[Likith D]

Just like you use the superposition principle to find the net $E$ field for a system of charges. You can superimpose a small disk on a sphere.

The Electric field due to a spherical shell at a distance $\delta$ away from the shell is $$ E=\frac {Q} {4 \pi \epsilon (R+\delta)^2} $$

And for a disk Electric field at a distance $\delta$ away is $$ E = \frac {\sigma} {2\epsilon} (1 - \frac {\delta} { \sqrt{ \delta^2+r^2 }}) $$ r:radius of disk(small)

Now if you place a disk with $-\sigma$ charge density at the surface of the shell you get the same field as that due to a shell with a hole.

$$ E=\frac {Q} {4 \pi \epsilon (R+\delta)^2} - \frac {\sigma} {2\epsilon} (1 - \frac {\delta} { \sqrt{ \delta^2+r^2 }})$$ where $\sigma = \frac {Q} {4 \pi R^2}$

For $\delta = 0$ you get $E = \frac {\sigma} {2 \epsilon}$ I think this should work even when $\delta =/= 0$

The same answer we get even when approaching from the inside of the shell.
I find it very beautiful.
I am also very interested to learn such methods, and would be glad to receive any textbook suggestion regarding the same.
Thanks a lot!