Electric potential equals the negative area under the graph

In summary, when discussing potential and electric field, the potential difference is equal to the negative of the area under the graph of E vs distance. This is defined as the negative integral of force times distance or V(initial) - area under the curve. This negative value is a result of the conservation of energy, as the force takes work to transform particular energy into kinetic energy. Therefore, potential is defined as the negative of the work done by the field.
  • #1
polaris90
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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?
 
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  • #2
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.
 
  • #3
polaris90 said:
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.
 
  • #4
thanks everyone
 
  • #5


I can explain the concept of electric potential and why it is defined as the negative area under the graph.

First, let's define electric potential. It is the amount of work required to move a unit positive charge from one point to another in an electric field. It is measured in volts (V).

Now, let's consider the graph of electric field (E) vs distance (d). The area under this graph represents the work done by the electric field on a unit positive charge. This work is equal to the change in potential energy of the charge.

In order to calculate the change in potential energy, we need to integrate the force (F) times the displacement (ds) over the distance. This is represented as V = ∫Fds.

However, the direction of the force and displacement may not always be the same. In fact, in most cases, the force and displacement will be in opposite directions. This means that the work done by the electric field will be negative.

In order to account for this negative work, we take the negative of the area under the graph of E vs d, which is equal to the negative integral of Fds. This is why the formula for electric potential is often written as V = -∫Eds.

To address your question about why it is not the final potential minus the initial potential, let's consider a simple example. Imagine a charge moving from a point A to a point B in an electric field. The initial potential at point A is V(A) and the final potential at point B is V(B).

If we were to calculate the potential difference as V(B) - V(A), it would give us a positive value. However, the work done by the electric field is negative, as it is in the opposite direction of the displacement. This is why we take the negative of the area under the graph, which gives us the correct value for the potential difference.

In summary, the negative area under the graph of E vs d represents the negative work done by the electric field, which is equal to the change in potential energy of a unit positive charge. This is why it is defined as the negative of the integral of Fds. I hope this explanation helps to clarify the concept of electric potential.
 

1. What is electric potential?

Electric potential is a physical quantity that represents the amount of electric potential energy per unit charge at a given point in an electric field.

2. What is the relationship between electric potential and the negative area under the graph?

The negative area under the graph represents the amount of work that must be done to move a unit charge from one point to another in an electric field. This work is equal to the change in electric potential, and since work is a negative quantity in this case, the electric potential is equal to the negative area under the graph.

3. How is electric potential measured?

Electric potential is measured in units of volts (V) using a device called a voltmeter. It can also be calculated by dividing the work required to move a charge between two points by the amount of charge being moved.

4. What is the significance of the negative sign in the equation for electric potential?

The negative sign in the equation represents the direction of the electric field. It indicates that the electric field is directed from the point of higher potential to the point of lower potential.

5. How is the concept of electric potential used in practical applications?

The concept of electric potential is used in many practical applications, such as in designing electrical circuits, measuring the strength of electric fields, and understanding the behavior of charged particles in different environments. It is also important in understanding the principles of electrostatics and electromagnetism.

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