Today we learned that in order for an object to move from one distance to another distance above the earth's surface, it needs a certain amount of force which increases its gravitational potential energy (work) to raise a mass from d1 to d2.
Electric potential energy is the same thing as when you place a positive charge between parallel plates. The charge is attracted to the negative plate but in order for the plate to move further away a force must be applied upward. Making a second height other than the first. The area of the shaded region between the two heights reps the increase in electric potential energy.
Calculating Electric potential energy.
PE(1)/(force*height1)=(Force)(distance of charge from negative plate)
=?J (amount of electric potential energy the charge has from the -ve plate)

Now you calculate the PE of charge at a second height
PE(2)=(Force)(second distance of charge from negative plate)
=?J(amount of electric potential energy the charge has from the -ve plate at a second distance)

Now to find the total PE b/w the two positions YOU SUBTRACT PE(1) from PE(2). Now the answer doesnt always have to be positive. Only if point B is further away from the negative plate. If point B is closer than it has a loss of PE.

This can be found from the shaded region of the two distances by multiplying the force of both(only if the same force is applied) by the two distances.


So PE is the work that is done in order to move a charge from one place to another.

ELECTRIC POTENTIAL
Electric potential and PE are very different. Electric potential is the potential energy that a charge has at a given point. Which can be found by this equation ( energy,J)/(charge,C)
The quantity is called a volt.

Potential Difference
To calculate the potential difference of a charge once it has moved to another point you must calculate the Electric potential of the first point V(1)=( energy,J)/(charge,C) and the same for the second point V(2)=( energy,J)/(charge,C). Doing the same thing as in Electric potential energy the Voltage of the first point from the voltage of the second point.

DeltaV=V(2)-V(1)


Now the amount of voltage that has changed in the positions of the charge is EQUAL to the amount of work needed to move the charge from one position to the next.

Now we were given a fun worksheet in which we went over in class and a second problem solving sheet which to be finished by tomorrow.