# When an object is dropped on earth, the work done by the gravitational force is equal to the change in gravitational potential energy: W = mgh initial.

## Presentation on theme: "When an object is dropped on earth, the work done by the gravitational force is equal to the change in gravitational potential energy: W = mgh initial."— Presentation transcript:

When an object is dropped on earth, the work done by the gravitational force is equal to the change in gravitational potential energy: W = mgh initial - mgh final

Similarly, when a charge is moved in an electric field, the work done on the charge by the field is equal to the change in electrical potential energy: W = EPE initial - EPE final.

The electric force on the charge in the field is F = q 0 E. It is useful to express this work per-unit-charge, by dividing both sides of the previous equation by q 0. W/q 0 = EPE initial /q 0 - EPE final /q 0

The value EPE/q 0 is called electric potential or the potential, or potential difference.

The electric potential V at a given point is the electric potential energy EPE of a small test charge q 0 situated at that point divided by the charge itself: V = EPE/ q 0 The unit is the joule/coulomb = volt (V).

The electric potential difference between two points is: V final - V initial = EPE final /q 0 - EPE initial /q 0 = -W/q 0 or, ∆V = ∆(EPE)/q 0 = -W/q 0

We only measure the differences of electrical potential V and electrical potential energy EPE, not their absolute amount.

Ex. 1 - The work done by the electrical force on a test charge (q 0 = +2.0 x 10 -6 C) as it moves from A to B is W AB = +5.0 x 10 -5 J. (a) Find the difference ∆(EPE) between these points. (b) Determine the potential difference, ∆V, between the points.

A positive charge accelerates from a region of higher electric potential energy (or higher potential) toward a region of lower electric potential energy (or lower potential).

A negative charge accelerates from a region of lower potential toward a region of higher potential.

Ex. 2 - Three points, A, B, C, are located on a horizontal line. A positive test charge is released from rest at point A and accelerates toward point B. Upon reaching B, the test charge continues to accelerate toward C. Assuming that only motion along the line is possible, what will a negative test charge do when it is released from rest at B?

An automobile battery has a potential that is 12 V higher at the positive terminal than at the negative terminal, V+ - V- = 12 V. The battery takes the positive charge that reaches the negative terminal and moves it to the positive terminal through the battery.

This requires work to be done to the charge by the battery; this energy comes from the battery’s reserve of chemical energy.

Ex. 3 - Determine the number of particles, each carrying a charge of 1.60 x 10 -19 C, that pass between the terminals of a 12-V car battery when a 60.0-W headlight burns for one hour.

The word “volt” is also used in the context of an energy unit called the electron volt. One electron volt is the change in potential energy of an electron (q 0 = 1.60 x 10 -19 C) when the electron moves through a potential difference of one volt.

This change in potential energy is q 0 ∆V = (1.60 x 10 -19 C)x(1.00 V) = 1.60 x 10 -19 J. So, 1 eV = 1.60 x 10 -19 J. One MeV = 10 +6 electron volts. One GeV = 10 +9 electron volts.

The total energy of an object includes all forms of kinetic and potential energy. E total = 1/2mv 2 + 1/2I  2 + mgh + 1/2kx 2 + EPE

Ex. 4 - (a) A particle has a mass of m = 1.8 x 10 -5 kg and a positive charge of q 0 = +3.0 x 10 -5 C. It is released from rest at point A and accelerates horizontally until it reaches point B. Angular speed  is zero. The only force acting on the particle is an electric force, and the electric potential at A is 25 volts greater than that at B. (V A - V B = 25 V) What is the speed v B of the particle when it reaches point B?

(b) If the same particle had a negative charge and were released from rest at B, what would be its speed v A at A?

When a positive test charge +q 0 is moved away from a positive point charge +q, work is done by the force between the charges. The magnitude of this force is given by Coulomb’s law: F = kq 0 q/r 2.

Since the distance from the charge +q varies, so does the force F. Therefore, the work is not simply F x d. Integral calculus can be used to find the work W AB when the particle +q 0 is moved from point A to point B.

W AB = kqq 0 /r A - kqq 0 /r B Therefore: V B - V A = kq/r B - kq/r A

At an infinite distance r B, V B and kq/r B become zero. The equation becomes: V = kq/r.

This V = kq/r is the potential of a point charge compared to the potential at infinity. This V has no value in the absolute sense, it refers to the difference from the potential an infinite distance away.

When q is positive, V is also positive, indicating that a positive charge raises the potential around it above zero. Conversely, a negative q decreases the potential below the zero reference value.

Ex. 5 - Using a zero reference potential at infinity, determine the amount by which a point charge of 4.0 x 10 -8 C alters the electric potential at a spot 1.2 m away when the charge is (a) positive and (b) negative.

When two or more charges are present, the potential due to all the charges is obtained by adding together the individual potentials.

Ex. 6 - Charges of +8.0 x 10 -9 C and -8.0 x 10 -9 C are separated by a distance of 0.80 m. What is the total electric potential at (a) a point exactly halfway between the two charges (0.40 m from each) and (b) between the two charges, 0.20 m from the positive charge and 0.60 m from the negative.

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