物理 energy of electron, radius

以下是我不明白的問題:

1. 為甚麼 electric potential energy 是 negative?我在網上找過資料,它們說 electric potential energy at infinity 是 definded 0,但是為甚麼要這樣定義?

2. 書中有一條問題:

A negatively charged particle A orbits around a fixed positively charged particle B.

The charges of A and B are -Qa and Qb respectively, and their masses are Ma and Mb. Given that the radius of the orbit of A is r

a) Find the Kinetic energy of A

ANS : QaQb/ (8πεr)

b) What happens to A if its Kinetic energy decreases.

Ans: By the equation in (a), when K.E. decreases, r increases.i.e. the radius of the orbit of A increases.

But,

一開始 electric force = 向心力

但假如KE decrease, electric force > 所需的向心力,所以應該會 spiral inwards

r 應該會 decrease

另外,Total energy of A = - QaQb/ (8πεr)

當KE decreases, Total energy decreases,所以 r 應該會decrease

這些推論是否正確?假如正確,那為甚麼與書的答案有矛盾?

希望各位幫忙!

更新:

Thanks For your answer !!!

But there's still something that I don't understand.

1. why the KE of particle A would increase when r decreases ?

Consider the circular motion, when radius decreases, the linear speed should decrease,which means KE should decrease.

更新 2:

2. Also, I understand that a decrease of KE is resulted by an increase of potential energy if total energy remains constant

更新 3:

But then, when KE decreases, PE increases, total energy remains constant

according to the equation of total energy = - QaQb/ (8πεr), r should remain unchanged. Hence the total enery seem not to be constant.

1 個解答

評分
  • 天同
    Lv 7
    5 年 前
    最佳解答

    1. This follows the definition in mechanics. Potential energy for repulsice force is +ve and for attractive force is -ve. The reason behind is that wok needs to be done when two +ve charges come near to each other, and -ve work is done (i.e. energy is being released) for charges of opposite signs come together.

    2. Should your argment is correct (i.e. r decreases), the KE of particle A would increase. This contradicts to what is asked in the question that what happen when KE of A decreases.

    This is simple mechanics, given the total energy remains constant, a decrease of KE results in an increase of potential energy (PE) and vice versa.

    2015-07-01 12:31:35 補充:

    Just to amend the last paragraph to read as:

    This is simple mechanics, given the total energy remains constant, a decrease of KE is resulted by an increase of potential energy (PE) and vice versa.

    2015-07-01 20:53:53 補充:

    Your suppl questions:

    1. Remember that particle A is under an attractive force (given by particle B). At a given orbit, there is a balance of electrical force and centripetal force, thus the KE and PE is fixed....

    2015-07-01 20:54:34 補充:

    (cont'd)...

    Should the speed of A decrease (by whatever means), particle A will be pulled down by the electrical force and hence increase its speed, leading to an increase of KE (this is a result of work done by the electrical attractive force).

    2015-07-01 21:02:08 補充:

    2. In a circular motion under an attractive field, the higher the orbit, the more is the total energy. But the same principle holds...

    2015-07-01 21:02:49 補充:

    (cont'd)...

    When particle A changes its orbit from a lower to a higher one, work needs to be done againt the attractive force between the two charges. This decreases the KE of the particle.

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