Boundary conditions example Infinite plane with charge density σ +σ Is the E field continuous across the plane ? No ! Is the electric potential V continuous across the plane ? Yes ! (energy is not created or destroyed) Conclusion is general Hints for 23.49 +q Assume the potential at infinity is zero rb -q ra A metal sphere with radius r is supported on an insulating stand in the center of a hollow metal spherical shell of radius b. There is a charge –q on the inner sphere and a charge +q on the outer shell. One approach: find the E fields and then the potentials. Match the boundary conditions Hints for 23.49 cont’d Make a Gaussian surface inside the sphere. What is E inside ? +q rb -q ra What is the electric potential for r<ra ? V=V0 we will determine this constant later. Hints for 23.49 (cont’d) Now make a spherical Gaussian surface at radius r +q r -q rb ra What is the charge enclosed ? What is the E field ? −q E= 2 4πε 0 r V (r ) = −q 4πε 0 r + V1 Hints for 23.49 (cont’d) Now make a spherical Gaussian surface at radius r outside. +q r rb -q ra What is the charge enclosed ? What is the E field ? What is the potential ? V (r ) = V2 = 0 Hints for 23.49 (cont’d) Now put all the electric potentials together. +q rb -q ra What happens at the boundaries ? V (r ) = V0 − q − q = ; = = + r r V V V a 0 1 r <r<r V ( r ) = + V1 πε 4 r 0 a 4πε 0 r −q + V1 r = rb ;V = 0 = r>r V (r ) = V2 = 0 4πε 0 rb r<r a a b b Hints for 23.49 (cont’d) r = rb ;V = 0 = −q + V1 4πε 0 rb −q + V1 r = ra ;V = V0 = 4πε 0 ra r<a a<r<b r>b V1 = q 4πε 0 rb −q q + V0 = 4πε 0 ra 4πε 0 rb Calculate Vab=Vb-Va V (r ) = V0 −q + V1 V = q ⎛ 1 − 1 ⎞ = V V (r ) = ⎜ ⎟ 0 ab 4πε 0 r 4πε 0 ⎝ ra rb ⎠ V (r ) = V2 = 0 Hints for 23.87 Nuclear Fission Q=96e U236 Q=46e 4π 3 V /2= Rdaughter 3 R=7.4 x 10-15 m Q=46e Half the volume and half the charge of the original nucleus What is the kinetic energy of each daughter nucleus when they are far apart ? Hints for 23.87 Q=96e R=7.4 x 10-15 m U236 What is the kinetic energy of each daughter nucleus when they are far apart ? Q=46e Rdaughter Rdaughte 2 k (46e) U= 2 Rdaughter U K= 2 Q=46e Numerical final result: 253 kilotons of TNT from 10 kg of uranium !! This is “electric potential energy”