In 1, Gaussian surface is a sphere with radius r, which is concentric with the sphere. From the symmetry, the field strength of each point on the surface is equal in radial direction, and from the Gauss theorem ∮ SEDS = (1/ε 0) ? ρ DV.

When r ≤ R, we get e1* 4 π R2 = (1/ε 0) ρ (4/3) π R3.

E 1=ρr/(3ε0)

When r > r, E2 * 4 π r 2 = (1/ε 0) ρ (4/3) π r 3 is obtained.

E2=ρR^3/(3ε0 r^2)

The potential energy φ (p) = ∫ (p: ∞) EDL at any point.

When r ≤ R, φ1(r) = ∫ (r: r) e1dr+∫ (r: ∞) e2dr = ∫ρr/(3ε 0) dr+∫ρr 3/(.

=ρ(3R^2-r^2)/(6ε0)

When r > r, φ 2 (r) = ∫ (r :∞) e2dr = ∫ρr 3/(3ε 0r 2) dr = ρ r 3/(3ε 0r).

2 With the center of the arc as the origin, put the arc directly above it (draw a picture yourself). Take a short section of ds on the arc, the angle between DS and Y axis is α, and the central angle of DS is ds dα, then ds=adα. This short charge dq=λds=λadα.

In which the electric field intensity at the center of excitation of wire density λ=q/aθ dq is de = dq/4 π ε 0a 2.

By the symmetry ex = 0e = ey = ∫ dey = ∫ de cos α = (λ/4 π ε 0a) ∫ (-θ/2: θ/2) cos α d α.

=(λ/4ω0a)* sinα=(λ/2ω0a)* sin(θ/2)

= q/(2 π ε 0a 2 θ) * sin (θ/2) direction downward.