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Transcript
Jackson 2.9 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: An insulated, spherical, conducting shell of radius a is in a uniform electric field E0. If the sphere is cut into two hemispheres by a plane perpendicular to the field, find the force required to prevent the hemispheres from separating (a) if the shell is uncharged; (b) if the total charge on the shell is Q. SOLUTION: (a) Align the z axis with the direction of the electric field. Find the potential outside a sphere at the origin in a uniform field by placing charges at z = -R and z = +R with charges +Q and -Q and letting R and Q approach infinity with Q/R2 constant. The response of the sphere can be represented by placing two image charges -Qa/R and +Qa/R in the sphere at -a2/R and +a2/R. The potential outside an uncharged conductor in a uniform field is therefore the potential of these four charges: Φ= Q 4 π ϵ0 [ 1 1 − 2 + 2 √ r +R +2 r R cos θ √ r +R −2 r Rcos θ 2 2 a /R √ r2 + 4 a r −2 a 2 cos θ 2 R R − a/R √ r2 + 4 a r +2 a 2 cos θ 2 R R In the limit R >> r, this becomes: Φ=−E 0 r cos θ+E 0 a3 cos θ r2 where E0 was recognized as 2Q/4πε0R2 The first term is just the potential due to the applied field in spherical coordinates. The second term is the potential of a perfect dipole. The sphere there has an induced charge distribution that acts as a perfect dipole. The electric field is therefore: E=−∇ Φ 3 ̂ a (2 cos θ r̂ +sin θ θ)] ̂ E= E 0 [cos θ r̂ −sin θ θ+ 3 r The electric field at the surface of the sphere is: E( r=a)=E 0 3 cos θ r̂ ] The charge distribution on the sphere's surface is found using: σ=ϵ 0 r̂⋅E(r =a) σ=3 ϵ0 E 0 cos θ If the sphere is now cut into hemispheres at the polar angle θ = π/2, the bottom hemisphere will feel a total force: F=∫ σ( x) E(x) d a We have to be careful and not include the force of the bottom hemisphere on itself. We do this by using the relation E= 2σϵ which gives us just the electric field at the surface of a conductor due to non-self contributions. Using this, we have: 0 F= 1 σ2 r̂ d a 2 ϵ0 ∫ F=k̂ 2π π a2 ∫ ∫ (3 ϵ0 E 0 cos θ)2 cos θ sin θ d θ d ϕ 2 ϵ0 0 π/ 2 F=k̂ 9 ϵ0 π E 0 a 2 2 π ∫ cos 3 θ sin θd θ π /2 9 2 2 F=− π ϵ 0 E 0 a k̂ 4 The force needed to keep the bottom hemisphere in place would therefore have to be equal and in the opposite direction: 9 2 2 F= π ϵ 0 E 0 a k̂ 4 Due the symmetry, the force needed to keep the other hemisphere in place would be equal and opposite. (b) If the sphere has a total charge of Q, it will just spread out uniformly on the sphere as an additional charge to the induced one. σ=3 ϵ0 E 0 cos θ+ Q 4 π a2 The total force on the bottom hemisphere will therefore be: F= 1 2 σ r̂ d a ∫ 2 ϵ0 F=k̂ 2 2π π a ∫ ∫ (3 ϵ0 E 0 cos θ+ 4 πQa 2 )(3 ϵ0 E 0 cos θ+ 4 πQa2 )cos θ sin θ d θ d ϕ 2 ϵ0 0 π/ 2 2π π F= a2 ̂ Q Q2 k ∫ ∫ (9 ϵ20 E 20 cos 2 θ+6 ϵ 0 E 0 cos θ + )cos θsin θ d θ d ϕ 2 ϵ0 0 π/ 2 4 π a 2 16 π 2 a 4 The first term represents the force on the induced charges due the external field and the field from the induced charges. The second term represents the force on the charge Q due the external field. The third term represent the force Q on itself. Note that the force of the external field on the point-charge-like charge Q will just tend to shift it and not separate it. Because we just want forces that will separate the two hemispheres, we must drop the middle term: 2π π a2 ̂ Q2 F= k ∫ ∫ (9 ϵ20 E 20 cos 2 θ+ ) cos θ sin θ d θ d ϕ 2 ϵ0 0 π/ 2 16 π2 a 4 F=−k̂ [ 9 Q2 π a 2 ϵ0 E 20+ 4 32 π ϵ 0 a 2 ] The force needed to keep the bottom hemisphere touching the upper sphere is therefore: F=k̂ [ 9 Q2 π a2 ϵ 0 E 20 + 4 32 π ϵ0 a 2 ]