AL Electric Potential Energy

👨🏻‍🏫 學術台其他 - 科學

[匿名]

2010-03-18 2:27 am

Proof of electric potential energy:
There is a point charge +Q and a test charge +q

Q<--------------------r------------------------>------q------------------------------∞
To move q from infinity to a distance r from Q,
- Electric Force on q = (1/ 4piε.)(Qq/r^2)
- Applied Force = - (1/ 4piε.)(Qq/r^2)
When we move the test charge q for a small distance dr ,
Work Done = - (1/ 4piε.)(Qq/r^2)dr
So, the total work done, i.e. the electric potential Energy
= (∞ to r)∫ - (1/ 4piε.)(Qq/r^2)dr
= (1/ 4piε.)(Qq/r)

However,
Since dr and applied force are in same direction (take right hand side is +ve)
why the small distance is dr , but not -dr??

更新1:

To physics8801 : for increment of vector.. q is moving towards Q x-------------------r----------------><---dr--- x---------------------r+dr--------------------> The increment seems to be -dr....>.

更新2:

from r + dr to r

回答 (2)

AAA

2010-03-19 1:12 am

✔ 最佳答案

Actually, the potential is measured as the work done of q moving towards Q. Hence, the force is pointing from Q to q. Taking right as positive, applied force is positive. Then if you take displacement as -dr, the result is the same as the above proof.
Another way to think of it is to neglect the positive and negative sign of dr and F (regard them as vectors), then use the definition of work done:
W = Fscos(theta)
Take theta as 180 degrees, the negative sign will appear too.

參考: Myself

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天同

2010-03-18 3:32 am

In the derivation, you have already asumed that the charge +Q is located at the origin. The position vector of the the test charge +q is measured from the origin. That is, the position vector is pointing from +Q to +q. Hnece, dr, which is an increment of vector r, is in the same direction pointing away from the origin.

Whether is the movement of the test charge +q is positive or negative is taken by the limit of intergration, i.e. from r to infinity, or from infin ity to r.

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