Alternatively, this problem can be solved by finding the solution to Poisson’s (Laplace’s) equation for r < R ( r > R ). In this method, we would find the solution to the differential equation ∇ 2 ϕ ( r ) = - ρ ( r ) /ε 0 . The electric field would then computed using E = -∇ ϕ . (b) The electric field and the potential are sketched in the figure. 1
(c) The stable equilibrium position is at the origin r = 0, at which the force due to the field of the charged sphere vanishes. (d) In equilibrium the total force vanishes F sphere + F ext = 0 . or qE ( r )ˆ r + qE 0 ˆ x = 0 . Using the expression for E ( r ) from part (a) inside the sphere, we find for the new equi- librium position d = E 0 4 πε 0 R 3 q ˆ x . We can also minimize the total potential energy obtained by adding - E 0 x to the elec- trostatic potential inside the sphere. (e) Outside the sphere, we can take its charge - q to be at the origin. Hence, the superpo- sition this negative charge with the positive point charge (+ q ) gives the system a dipole moment of p = q d = (4 π ε 0 R 3 ) E 0 ˆ x . 2
(f) Comparing the expression in (e) with with p = α E , we find α = 4 π ε 0 R 3 . Interestingly, the polarizability depends only on the radius of our classical atom; the larger the radius the larger the polarizability, meaning that our classical atom exhibits a larger response to a given applied field. 3
Problem 2: Electromagnetism II (a) Using Biot-Savart Law (in SI units): d ~ B = μ 0 I d ~ l × ˆ r 4 πr 2 , where d ~ l is the differential directed length of the current, and r is the distance from the differential current to the observation point.
You've reached the end of your free preview.
Want to read all 5 pages?
- Winter '16