P22_014.PDF

(67 KB) Pobierz
Chapter 22 - 22.14
14. (a) We choose the coordinate axes as shown in the diagram below. For ease of presentation (of the
computations below) we assume Q> 0and q< 0 (although the final result does not depend on
this particular choice). The repulsive force between the diagonallyopposite Q ’s is along our (tilted)
x axis. The attractive force between each pair of Q and q is along the sides ( o f length a ). In our
itself is therefore of length 2 d = a 2.
. . . . . . . . . . . . . . . . . . . . . .
x
q
a
Q
d
a
d
Q
q
Since the angle between each attractive force and the x axis is 45 (note: cos 45 =1 / 2) , then the
net force on Q is
F x =
1
4 πε 0
( Q )( Q )
(2 d ) 2
2 (
|
q
|
)( Q )
a 2
cos45
Q 2
2 a 2
=
1
4 πε 0
2 |
q
Q
1
2
a 2
which (upon requiring F x =0)leadsto
|
q
|
= Q/ 2 2or q =
Q
2 2 .
(b) The net force on q , examined along the y axis is
F y =
1
4 πε 0
q 2
(2 d ) 2
2 ( | q | )( Q )
a 2
sin45
q 2
2 a 2
=
1
4 πε 0
2 |
q
Q
1
2
a 2
2 Q 2 which is inconsistent with the result of part (a).
Thus, we are unable to construct an equilibrium configuration with this geometry, where the only
forces acting are given byEq. 22-1.
drawing, the distance between the c enter to the corner is d ,where d = a/ 2, and the diagonal
y
which (if we demand F y =0)leadsto q =
643283672.001.png
Zgłoś jeśli naruszono regulamin