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Question about electric field strength...Please help!!!!?
So this equation is like
Vf - Vi = (Fd)/q = -Ed (http://hyperphysics.phy-astr.gsu.edu/hbase/electri... the diagram is on this link
Why is the work done taken as negative ie "-E"?
What would happen if the charge were to be moving from +ve to -ve terminal? I mean at that time no external force would need to do any work right? The electric field will drag the point charge to the negative terminal, right?
Can you explain the entire derivation and the concept in a bit simpler way....
1 Answer
- 9 years agoFavourite answer
This concept is kind of confusing to many undergraduate students. Here is the reason for the involevement of a negative sign :
You see when we move a +ve charge from region of low electric potential to region of high electric potential, the potential energy concerned with the +ve charge increases.
So, if we move a +ve charge from -ve potential to +ve potential, actually, the potential energy of +ve charge increases.
So, Vf - Vi = +ve quantity
And now, the "Ed" specifically mentioned in the link is not the work done by us but instead it is the work done by Electric field. If we do a +ve work say +Ed, the work done by Electric field correspoding to our work will be -Ed.
So, Vf - Vi = (Work done by us) = -(Work done by Electric Field) = -(Ed)
This Ed is a -ve quantity
=> -Ed is a +ve quantity.
So,
Vf - Vi = +ve quantity, which is true.