How to solve the field strength on a uniformly charged spherical surface?

There are two concepts that need to be clarified! Spherical shells and spherical charge distributions are different!

1。The charge distribution of the sphere is similar to that of the whole sphere, that is, the sphere is composed of countless charged charges, and the amount of charge surrounded by the Gaussian surface is calculated by the volume uniformity, that is, the percentage of the volume surrounded by the Gaussian surface to the volume of the original sphere multiplied by the total charge of the original sphere!

2。The spherical shell charge is only distributed on the surface of the sphere, and the reverie is very thin, and any Gaussian surface smaller than the radius of the sphere has no charge in it, so the field strength is zero.

Of course, if the spherical shell has thickness, it is not difficult to solve, that is, the intermediate state of the above two types, let the inner diameter of 1 and the outer diameter of 2 spherical shell.

The field strength of the Gaussian surface greater than 2 can be directly substituted into the known total charge according to the formula.

The Gaussian surface is between 1 and 2, and it can be calculated using the method of 1 above.

The field strength charge of the Gaussian surface less than 1 is zero, so the field strength is also zero!

The electric field lines in reverie are easy to understand, because the charges are uniformly distributed, whether positive or negative, and only one type of homogeneous charge can exist. Therefore, for any infinitesimal point of the sphere, that is, a microscopic charge, an isotropic charge can always be found at a relative diameter distance, so the electric field formed by the charge cancels out everywhere according to the superposition principle! Hence the spherical shell inner field strength is zero.

It can be proved by the charge surface distribution integral of the spherical shell!

Hope it helps!!

You're asking about the university, using Gauss's theorem.

Addendum: The Gaussian surface of the spherical shell and the sphere should be the same.

Gaussian, taking the problem as an example, if the sphere with a penalty is taken as the Gaussian surface, then the flux is e*s then q 0=e s

q is the amount of electricity inside the Gaussian surface.

Reason: When it is charged, the same charge will be as far away as possible because the same charge repels each otherEvenly distributed barricadesOn the outer surface, the internal field strength should be 0 when equilibrized.

The strength of the electric field at a point in the electric field.

The direction of the electric field force exerted at the point of the tentative charge (positive charge) can be used.

the direction of the electric field; The strength of the electric field can be determined by the ratio of the force on the test charge to the charge charge at the test point.

Introduce. The electric field strength follows the principle of superposition of field strengths, that is, the total field strength in space is equal to the vector sum of the field strengths when each electric field exists alone, that is, the principle of superposition of the field strength of air combustion is an experimental law, which shows that each electric field is acting independently and is not affected by the presence of other electric fields.

The above described description applies to both electrostatic fields and general electric fields with spin electric fields or both. The superposition of electric field strengths follows the parallelogram rule of vector synthesis.

A physical quantity that describes the strength or weakness of an electric field, and a physical quantity that describes the nature of the force of an electric field. The magnitude of the electric field strength depends on the electric field itself, or rather on the charge that excites the electric field, and has nothing to do with the charged charge in the electric field.

The charge is q and the radius is r. The internal field strength e=0 outside the sphere, the equivalent charge at the center of the sphere is a little charge e=kq r 2 r>r, and the electric potential is equal to the outside of the sphere, and the equivalent charge at the center of the sphere is a little charge =kq r. If it is a uniformly charged sphere, the result is the same as that of a spherical shell.

The Gaussian sphere is made at a distance from the center of the sphere r, and the electric flux on the sphere is (4 3 r δ because the field strength is evenly distributed, so the size of the field strength can be directly divided by the area 4 r.

It is necessary to find the potential distribution inside and outside the sphere separately, the first one is to find the field strength distribution, and according to du=edr, the potential is found integrally. The second is based on the principle of potential superposition, if it is outside the sphere, it is directly regarded as the point charge at the center of the sphere, if it is inside the sphere, the ball needs to be divided into two parts, and the potential generated by the internal part is solved in the same way, and the external part needs to be integrated.

The field strength of a point close to the spherical shell outside the spherical shell is kq r 2, the field strength of a point close to the spherical shell inside the spherical shell is 0, the field strength generated at this point outside the spherical shell is e1, and the field strength generated by this point at a very close place is e2, then e1+e2=kq r 2, e1-e2=0

So the field strength of the spherical shell against this point is kq 2r 2.

If we use Gauss's theorem, let the area of this point (a very small surface) be s, make a very small cylinder containing this surface, and the base area is equal to it, because the cylinder is small, so there are no electric field lines passing through the side. Let the areal density be =q 4 r 2, then 4k s=e*2s, and get e=kq 2r 2

According to Gauss's law of electric field, the area fraction of the electric field strength on any closed surface in space is equal to the ratio of the sum of the charges in the surface to the space permittivity. Namely:

eds=∫(ρ/ε)dv

Now we can assume the simplest case, where there is only one charged metal ball in space (the charges are evenly distributed over the surface). Since the charged spherical surface is symmetrical, its charge distribution is also symmetrical with respect to the center of the sphere, so its electric field is also spherically symmetrical.

Under the convenience provided by this symmetry, we may wish to choose the integrating surface as a sphere concentric with the charged sphere, and the direction of the electric field at each point on the integrating sphere is perpendicular to the spherical surface, and the magnitude of the electric field strength is equal everywhere. Therefore, we can simplify the above integral formula to:

4πr²e=∫(ρ/ε)dv

The right side of the equation represents the amount of charge contained within the integrating sphere.

It is not difficult to find that when the integrating sphere gradually shrinks from infinity, the electric field changes uniformly with the change of the spherical diameter, until when crossing the charged sphere, the electric field e suddenly changes from q 4 r to 0.

The root cause of this phenomenon is a sudden change in the amount of charge contained within the surface.

edr= (q 4 r)dr=q 4 r (the upper and lower limits of the integration are and r, respectively, and r is the radius of the spherical surface with points. ）

The field strength is greatest when the sphere is charged.

In the ball, the field strength at the center of the ball is 0, then the change from the center of the ball to the field strength of the sphere is an increase function outside the sphere, the field strength decreases with the increase of distance, and from the sphere to infinity is a decrease function from the increase function to the decrease function, of course, it is a mutation.

q/4πε0r

1.The charge q is evenly distributed.

On a sphere of radius r, the unit volume.

The charged charge of =q (4 r is the key of the old Bi key 3), the radius of r(r) always contains the whole charged sphere, so the total charge q=q;

2.For a uniform several volts charged gold cylinder, the charge values are distributed on the surface, and within the cylinder (r