Electric field The problem: Given a thin sphere, radius R, charge Q (divided evenly across the sphere), calculate the electric field inside and outside of the sphere: 1. using direct integration (the Coulomb’s law). 2. using the Gauss’s law. Using the result, calculate the electric field of a homogeneously charged solid sphere, radius R, charge Q. The solution: 1.According to the Coulomb’s law ~0 ~ = kdq(~r − r ) dE |~r − r~0 |3 In our case: Q Q dq = = σ = da A 4πR2 2 da = R sin θdθdφ Q sin θ dq = dθdφ 4π For simplicity let us choose (1) (2) (3) (4) ~r = (0, 0, z) (5) (any other point can be reached by rotating the coordinates) and r~0 = (R sin θ cos φ, R sin θ sin φ, R cos θ) (6) then |~r − r~0 | = (R2 + z 2 − 2Rz cos θ)1/2 (7) Therefore, E= kQ 4π Z π Z sin θdθ 0 2π dφ 0 1 (−R sin θ cos φ, −R sin θ sin φ, z − R cos θ) (8) (R2 + z 2 − 2Rz cos θ)3/2 Ex and Ey both equal 0 (since sin φ and cos φ are both periodic for 2π). Z kq π sin θ(z − R cos θ) Ez = dθ 2 0 (R2 + z 2 − 2Rz cos θ)3/2 substituting cos θ for t we get: Z kQ 1 z − Rt Ez = dt 2 2 −1 (R + z 2 − 2Rzt)3/2 Z 1 ∂ dt kQ = · (−1) 2 ∂z −1 (R2 + z 2 − 2Rzt)1/2 kQ ∂ 1 2 2 1/2 1 = (R + z − 2Rzt) |−1 2 ∂z Rz 1 kQ ∂ = (|R − z| − |R + z|) 2 ∂z Rz ( kq/z 2 , |z| > R Ez = 0, |z| < R 1 (9) (10) (11) (12) (13) (14) 2. According to the Gauss’s law: Z ~ · d~a = 4πkq Φ= E (15) Let us find the electric field at the point r1 (spherical coordinates). We build a sphere with a radius r1 . Due to the symmetry, the magnitude of the field on the sphere is the same, and in the r̂ direction. Then ~ ·A ~ = E · 4πr2 = 4πkq Φ=E 1 (16) where Q being the charge inside the sphere r1 . If r1 > R then all of charge is in the sphere, and we get : E= kQ r2 (17) and if r1 < R, then no charge is inside the sphere r1 , and so: E=0 (18) which is, of course, the same result as we got before. 3. Let us think of the solid sphere as a collection of thin spheres. Each thin sphere at the radius r0 < R and of thickness dr0 is charged with the charge dq = ρdV = Q dr0 4/3πR3 (19) For r > R each thin sphere induces dEr = kdq . Therefore, Er = kQ . r2 r2 For r < R one has to consider only the thin spheres with the radii r0 < r (since the outer spheres 3 contribute zero). Therefore, the total contributing charge is q = Q Rr 3 , and the field is Er = kQ r. R3 Summarising, ( kQ r, r < R R3 Er = kQ (20) , r>R r2 2