E field 問題

We know that F=Eq, when a charged particle moved along an electric field line. There is change in voltage, becacuse V=kq/r, it will have a voltage on the point charge. The first sentence is wrong. V=W/q=Fd/q=Eqd/q=Ed, The electrical potential energy of the charge undergoes the maximum unchange in magnitude, because we see V=Ed, doesn't concern how magnitude, only care the E field, so it will change the E field

2007-05-21 11:08:13 補充：
If the E field is between a plate, we should think F _|_ E, if it is _|_, it will be uniform. if not, it will have a angel. The voltage will be the same becsue the same distance.

2007-05-21 11:08:29 補充：
if the charge moved along an electric field line, the voltage will change because the e field is changing, we remember that V= E*ds

第1句
there is no change in voltage.
This statement is wrong.
An electric field line joins place of higher potential to that of lower potential. The potential decreases as one goes along the field line.

第2句:

the electrical potential energy of the charge undergoes the maximum change in magnitude.
This statement is correct.
The fall in (electrical) potential energy along the field line is max since field lines are always perpendicular to equipotential lines.
A charge that moves along the field line indicates that it goes in a direction perpendicular to the equipotentials, which will traverse the shortest distance between any two equipotentials. The drop in potential is therefore maximum.
On the other hand, if the charge moves at some angle to the field lines, then for the same distance travelled, it will traverse through less equipotential lines than one that goes along the field lines.

2個都錯.

之後你補充個part要註明其移動方向才有決定.