Answers for Assignment 4

      (a) V(x) = [ k q / | x-a | ] + [ k q / | x+a | ] , and you really need the absolute values.

      (b) [graph]

      (c) When V is a minimum, E(x)= - dV/dx = 0

20.
      (a) V = { k q }{ [ 2 / sqrt(x² + a² ) ] + [ 1 / | x - a | ] }

      (b) E_x (x) = kq [(2x)/(x² + a² )^3/2 + 1/(x-a)² ] ,      (x > a)

            E_x (x) = kq [ (2x)/(x² + a² )^3/2 - 1/(x-a)² ] ,      (x < a)

where k = 1 / (4 pi epsilon₀ )

      (c) The eventual kinetic energy of a fourth charge released from the origin will be

KE_f = q V(0) = 3 k q /a

21.

Using k as in the previous problem,

      V = k (Q / L ) ln [ (1 + L/d ) ]

22.

Again using k = 1 / (4 pi epsilon₀ ),

KE₁ = k [ q² /L ] [ 2 + 1 / sqrt(2) ]

KE₂ = k [ q² /L ] [ 1 + 1 / sqrt(2) ]

KE₃ = k q² /L

KE₄ = 0

KE_total = k [ q² /L ] [ 4 + sqrt(2) ]

23.

E (x, y, z) = - 4 x 1_x - z 1_y - y 1_z volt/m²

24.

Q_a = - Q (a/b) is the charge on the inner shell

25.

      (a) Answer given in the problem.

       (b) E_inside = [ pi rho₀ r² ] / [ 4 pi epsilon₀ R ]

             E_outside = [ pi rho₀ R³ ] / [ 4 pi epsilon₀ r² ]

       (c) V_inside = pi rho₀ (4 R³ - r³ ) / (12 pi epsilon₀ R )

             V_outside = pi rho₀ R³ / ( 4 pi epsilon₀ r )

Last revised 1998/03/09