(9分)将电量q 1 =+1.0×10 -8 C的点电荷,在A点时所受电场力大小是2.0×10 -5 N。将它从零电势O

1个回答

  • 解题思路:

    (1)由电场强度的定义式可得:

    A

    点处电场强度大小:

    E

    =

    =

    2000

    N

    /

    C

    (2)在电场中将电量为

    1.0

    ×

    10

    6

    C

    的点电荷从

    O

    移到

    A

    ,克服电场力做功

    2.0

    ×

    10

    6

    J

    ,则

    O

    A

    间电势差

    U

    O

    A

    =

    =

    200

    V

    ,则A.

    O

    间电势差

    U

    A

    O

    =

    200

    V

    (3)由

    O

    A

    :

    U

    O

    A

    =

    φ

    O

    φ

    A

    =

    "

    200

    V

    "

    W

    O

    A

    =

    U

    O

    A

    q

    2

    =

    4.0

    ×

    10

    6

    J

    (1)(2)(3)

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