解答 30.38. Solve: (a) (b) Equation 30.3 gives the potential ...

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解答


30.38.

Solve:

(a)


(b)

Equation 30.3 gives the potential difference between two points in space:


Taking
V
(
x
i
)


0 V at
x
i



0 m, and replacing
x
f

with simply
x
,
V
(
x
)



(2500
x
2
) V.

(c)

A graph of
V

versus
x

over the region


1 m


x



1 m is shown in part (a).

Assess:

As it must be, we have



30.40.

Model:

The electric field is the negative of the slope of the graph of the potential function.

Visualize:

Please refer to Figure P30.40.

Solve:

We have



dV



E
x
dx

For
x



1 cm,
E
x



1000 V/m. Integrating the above expression,

V





(1000 V/m)
x



C

Because
V



0 V at
x



0.03 m,
C



(1000 V/m)(0.03 m)


30 V. Thus,

V





(1000 V/m)
x



30 V

(
x



1 cm)

This means
V

at
x



0.02 m is 10 V and
V

at
x



0.01 m is 20 V. For
x

< 1 cm,
E
x



(10
5

V/m)
x
. Therefore,
dV





(10
5

V/m)
x dx
. Integrating again,


(
x

< 1 cm)

Because
V



20 V at
x



0.01 m,
C



25 V. At
x



0 m,
V



25 V.

30.4
1.

Model:

Assume the electrodes form parallel
-
plate capacitors with a uniform electric field
between the plates.

Visualize:


Please refer to Figure P30.41. The three metal electrodes serve as plates for two capacitors. On the middle
electrode, half the ch
arge is located on the left face and half on the right face, thus forming two capacitors.
Each plate of the two capacitors carries a charge of

50 nC.

Solve:

(a)

In the space 0 cm <
x

< 1 cm, the electric field points to the left and its magnitude is


In the region 1 cm


x



2 cm,

because in electrostatics the inside of a conductor has no free charge.
The electric field in the region 2 cm <
x

< 3 cm points to the right and has the same magnitude as the

electric
field in the region 0 cm


<
x

< 1 cm.

(b)

The potential difference between two points in space with a uniform electric field is


Assuming that the negative plate at
x



0 m is at zero potential (
V
i



0 V at
x
i



0 cm),
V
f



x
f
E
, or simply
V


xE
. Thus, the potential increases linearly with distance
x

from the negative plate in the region 0


x



1. At
x



1 cm, the potential is

V


xE



(1.0


10

2

m)(1.41


10
7

V/m)


1.41


10

5

V

The potential must be the same throughout the region 1 cm


x



2 cm. If this were not the case, we would
not have an electrostatic situation with the electric field
E



0 V/m. Using the previous reasoning, the
potential decreases linearly
in the region 2 cm <
x

< 3 cm.


33.27.

Model:

A magnetic field exerts a magnetic force on a moving charge.

Visualize:

Please refer to Figure EX33.27.

Solve:

(a)

The force is


(b)

The force is



33.61.

M
odel:

Electric and magnetic fields exert forces on a moving charge. The fields are uniform
throughout the region.

Visualize:

Please refer to Figure P33.61.

Solve:

(a)

We will first find the net force on the antiproton, and then find the net acceleration us
ing
Newton’s second law. The magnitudes of the electric and magnetic forces are


The directions of these two forces on the antiproton are opposite.

points
up

whereas, using the
right
-
hand rule,

points
down
. Hence,



(b)

If

were reversed, both

and

will point
up
. Thus,