Electric potential equals the negative area under the graph

[polaris90]

When taking about potential and electric field, potential difference is equal to the negative of the area under the graph of E vs distance? why is that. My book defines it as the negative integral of Force times ds or V(intitial) - area under the curve. I don't understand why it's negative. I see it's the initial minus the entire are which would give me a negative potential difference, but why isn't it the final minus initial?

 

[rcgldr]

It's by definition, potential is defined to be the negative of the work done by a field. It makes sense when you consider situations where total energy = potential energy + kinetic energy, such as an object being accelerated by a field (with no losses). For example, a positively charged object being accelerated from the positive plate of a capacitor to the negative plate in a vacuum (no drag); as the object accelerates away from the positive plate, it's kinetic energy increases and it's potential energy decreases, and total energy remains constant.

 

[zhangyang]

When taking about potential and electric field, potential difference is equal to the negative of the area under the graph of E vs distance? why is that. My book defines it as the negative integral of Force times ds or V(intitial) - area under the curve. I don't understand why it's negative. I see it's the initial minus the entire are which would give me a negative potential difference, but why isn't it the final minus initial?

The negetive is result from:
When a force take work,it transform the particular energy(such as electric inertial energy) into kinetic energy ,accoding to the conservation of energy, the particular energy is descreased.So f*s=-ΔE,so
ΔE=-∫f*s.

 

[polaris90]

thanks everyone