Unformatted text preview: ' 1 0 1 ( # 1 C1 # 1
C1 C1 *
(d ! a) 2 d ! a
!0 A !0 A d
(b) C #
d d !a
(c) As a E 0 , C E C0 . The metal slab has no effect if it is very thin. And as a E d , C E F . V # Q / C . V # Ey is
the potential difference between two points separated by a distance y parallel to a uniform electric field. When the
distance is very small, it takes a very large field and hence a large Q on the plates for a given potential difference.
Since Q # CV this corresponds to a very large C.
(a) IDENTIFY:! The conductor can be at some potential V, where V = 0 far from the conductor. This potential
depends on the charge Q on the conductor so we can define C = Q/V where C will not depend on V or Q.
(b) SET UP:! Use the expression for the potential at the surface of the sphere in the analysis in part (a).
EXECUTE:! For any point on a solid conducting sphere V # Q / 4, !0 R if V # 0 at r E F.
C# % 4, !0 R &
( # 4, !0 R
)Q* (c) C # 4, !0 R # 4, . 8.854 " 10!12 F/m / . 6.38 " 106 m / # 7.10 " 10!4 F # 710 $ F. 24.68. EVALUATE:! The capacitance of the earth is about seven times larger than the largest capacitances in this range. The
capacitance of the earth is quite small, in view of its large size.
IDENTIFY:! The electric field energy density is 1 !0 E 2 . For a capacitor U #
SET UP:! For a solid conducting sphere of radius R, E # 0 for r 1 R and E #
for r 5 R .
4, !0 r 2
EXECUTE:! (a) r 1 R : u # 1 !0 E 2 # 0.
(b) r 5 R : u # ! E # ! '
) 4, !0 r * 32, !0 r
20 2 F 1
20 (c) U # H udV # 4, H r 2udr #
R 24.69. F Q 2 dr
H r 2 # 8, !0 R .
8, !0 R Q2
(d) This energy is equal to 1
which is just the energy required to assemble all the charge into a spherical
2 4, !0 R
distribution. (Note that being aware of double counting gives the factor of 1/2 in front of the familiar potential energy
formula for a charge Q a distance R from another charge Q.)
EVALUATE:! (e) From Equation (24.9), U #
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.
- Fall '08