A sphere with a radius of R with a dotted body density distribution of P qr RRRR.

The charge of a spherical shell with radius r dq=p*4 r dr=(4q r 4)r dr.

Integral: Q=(4Qr 4)*R4 4=q.

This problem needs to cut the ball into an infinite number of thin slices, then cut the thin slices into an infinite number of rings, and then cut each ring into an infinite number of small dots, use the electric field formula e=k*q r2 to calculate the electric field of each point on a point outside the sphere, and then integrate to obtain the electric field of the ring to the point, then integrate to obtain the electric field of the sheet to the point, and finally integrate to obtain the electric field of the sphere to the point.

Since the sphere is symmetrical, the direction of the electric field should be radial, so only the electric field in the radial direction needs to be considered.

Gauss's theorem

Assuming that the charge is distributed in a curve or a straight rod, its line charge density is the charge density per unit length in coulombs.

Rice. Assuming that the charge is distributed on a plane or the surface of an object, the surface charge density is the charge density per unit area in coulomb meters2.

Suppose the charges are distributed in a three-dimensional space.

, then its bulk charge density is per unit volume.

The amount of electric charge in coulombs 3.

dq=p*4*π*r*r*r/3

q=∫(0~r)dq=4qr/15

The upper and lower limits of the points will not be played, so they are marked in parentheses, and the points are obtained from 0 to r).

Use Gauss's theorem es=q

A family of concentric spheres can be obtained by Gauss's theorem.

The charge of a spherical shell with radius r dq=p*4 r dr=(4q r 4)r dr integral: q = (4q r 4)*r 4 4=q or according to the Gaussian flux theorem: electric field flux along a closed du surface = enclosed charge dielectric constant;

Select a concentric sphere with a closed surface as radius r, r r, then 4 r * e = 4 3 * r * ; Yes e = r (3 )*1 r );

Select the point where the infinity is 0 potential, and the integral of e from r to 1 r is: 1 r, that is, the electric potential at the outer radius r of the sphere = 1 r;

substitution r=r; Ball surface potential = 1 r;

Find the potential at any point in the body = the potential of the surface of the sphere;

Solution: Let the total charged amount of the whole sphere with a radius r be used as an integral 5261q (total) = p*4 *r 41022dr (integral limit from 16530 to r) = kr 4. When r>=r, e=q(total) 4 * 0*r 2 (where 0 is the vacuum permittivity.)

r in the electric field caused by the continuous charge distribution is not a fixed value. In the actual calculation, the DQ cannot be directly integrated, and the element must be exchanged, and finally the DR must be commuted.

The surface area is 4 rr, and the electric charge at this radius is q=4 rr*, and the electric field strength.

e=kq/rr。

k is Coulomb. The constants in the theorem are different from k.

Therefore, the spatial field strength distribution e=k4 rr*kr rr=4 kkr to find the surface area of balls with different radii, multiply it by the density is the amount of charge, you can regard the electric field strength as the charge equivalent of the surface in the center, and find the point charge electric field.

The electric potential is equal to e r

The process of solving the problem is as follows:

Due to the uniform distribution of positive charges in the mold.

on the sphere, so the electric field strength.

There is spherical symmetry.

Let r be the straight-line distance from the center of the sphere to a certain point.

According to Gauss's theorem.

e=1 0 q (q is the charge of all charges contained within the Gaussian surface) For a sphere, e=e ds=4 r 2e

So when 1 0 q = 4 r 2e e= q ( 04 r 2)r r, the field point is not in the sphere, and the total charge q is the total charge contained in the charged body.

e=（4/3πr^3ρ）/04πr^2）=（r^3ρ）/3ε0r^2）

re=(4 3 r 3 ) 04 r 2) = (r ) 3 0) electric potential is equal to e r

Answer] Dust fiber: the radius is r, and the spherical shell with the thickness of Dr is micro-element then the micro-element charge dq=p0(1-r r)4 r dr so the sphere charge is q=4 p0 (r -r r)dr substituting the upper limit of the integral r the lower limit of 0 points to obtain q= r p0 3 then the field strength distribution of the imitation field of the ball is e=q 4 r =p0r 12 r (r