(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) Ex (x) = kq [(2x)/(x2 + a2 )3/2 + 1/(x-a)2 ] , (x > a)
Ex (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
KEf = 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
KEtotal = k [ q2 /L ] [ 4 + sqrt(2) ]
23.
E (x, y, z) = - 4 x 1x - z 1y - y 1z volt/m2
24.
Qa = - Q (a/b) is the charge on the inner shell
25.
(a) Answer given in the problem.
(b) Einside = [ pi rho0 r2 ] / [ 4 pi epsilon0 R ]
Eoutside = [ pi rho0 R3 ] / [ 4 pi epsilon0 r2 ]
(c) Vinside = pi rho0 (4 R3 - r3 ) / (12 pi epsilon0 R )
Voutside = pi rho0 R3 / ( 4 pi epsilon0 r )
Last revised 1998/03/09 |