Download 習題六 25.41. (a) The potential on the x axis is (b) The potential on

習題六
25.41. (a) The potential on the x axis is
𝑉=
𝑞
2𝑞
1
−
= 0 ⇔ 𝑥 = 𝑑, −𝑑
|𝑥| |𝑑 − 𝑥|
3
(b) The potential on the y axis is
𝑉=
𝑞
2𝑞
√3
−
=0⇔𝑦=±
𝑑
|𝑦| √𝑑2 + 𝑦 2
3
(c) If the charge is positive the potential of each charge is always positive, so there is no point at zero
potential except infinity
E (r  R/2)  0
E ( R/2  r  R) 
E (r  R) 
 
R3 
 r  2  radially outward
3 0 
8r 
7  R3
radially outward
24 0 r 2
25.45. (a)
(b) Ex 
V 500 V

 5.0 103 V/m (c) Here we treat the system as consisting of the parallel plates
x 0.10 m
and the electron. In that case
U  qV  (1.60  1019 C)(500 V)  8.0 10 17 J. (d) Now we treat
the electric field between the plates as being external, and we take the system to be the electron alone. In that
 V
case the work done by the electric force is equal to W  F E  x  qEx x  q  
 x

 x  qV . Clearly

the answer is the same in magnitude as the answer to (c). The only difference is that now we are discussing a
positive external work being done on the system. W  (1.60  1019 C)(500 V)  8.0 10 17 J. (e)
Whether electrostatic potential energy is converted into kinetic energy, or external work speeds up the
electron, either treatment tells us that the kinetic energy must be 8.0 1017 J.
25.59. We know Ex  
dV
, so E  B, and the electric field points in the  x direction.
dx