Bukhovtsev-et-al-Problems-in-Elementary-Physics

300
ANSWERS AND SOLUTIONS
.0 E
+
+
+
+ + +
tI+ - d
0.-- --.0
+
-(J + +Q +
+
+
+ :t
+ + +
+ + +
+ 1/ F
Fig. 414 Fig. 415
will move to the remote edges of the plate and tneir field may be. neglected.)
This distribution of the induced charges does not depend on the thickness
of the plate.
Let us place a charge -Q on the left of the plate and at the same
distance d. Obviously, the induced positive charges will be distributed on
the left side of the plate in the same manner as the negative charges on
the right side. The charge -Q placed on the left of the plate will not
cause a change in the electric field at the right of the plate. Thus, at the
right of the plate, the electric field induced by the charge +Q and the
induced negative charges coincides with the field produced by the charges
+Q and - Q and the charges induced on the surfaces of the plate (Fig. 414).
If the thickness of the plate is small as compared with d, the plate
may be regarded as infinitely thin and hence the field created by the induced
charges outside the plate is zero.
It has been shown that the field on the right of the plate produced by
the charge +Q and the induced negative charges is equal to the field
caused by the point charges +Q and -Q.
Since the intensity of the field caused by the induced negative charges
at the point where the charge +Q is equals the intensity of the field caused
by the point charge -Q at a distance of 2d from + Q, the sought force
of attraction will be F= ~: .
414. Since a and .b are much greater than c and d. it can be assumed
that the plate is infinitely large. Remembering that the intensity of a field
caused by several charges is equal to the sum of the intensities produced by
each of these charges, and using the results of the solutions of Problems
412 and 413, we can obtain the sought force:
Qt
2nQQ 1
F=----
ab 4d l
The plus sign corresponds to the force of repulsion and the minus sign to
the force of attraction.