topic 6.2 extended d – electric potential energy and p.d

Topic 6.2 ExtendedD – Electric potential energy and p.d.

In mechanics we began with vectors, and then found easier ways to solve problems - namely energy considerations.

In the last chapter we looked at electric vectors. Now we want to look at electric energy. This will simplify finding solutions to electrical problems.


Consider a mass in a gravitational field:

If we raise the mass from a height of r0 to a height of rf and release it, we know it will relinquish the potential energy we gave to it - converting it to kinetic energy.

GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - POINT MASS

rf

r0

The gravitational force is given by


rf

r0

FG = GmMr2

and the work done against gravity is the potential energy difference

UG = FG rf - FG r0

Gravitational Potential Energy Difference

(Point Mass)

=GmM

rf2

•rf -GmM

r02

•r0

UG =GmM

rf

-GmM

r0

GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - POINT MASS

An electric field has the same properties as a gravitational field.If we place a test charge q0 in an electric field we must do work on it in moving it from one place to another.


A

q0

q0

r0

rf

If we release the test charge it will relinquish the potential energy we gave to it, just as the mass did in the gravitational field.The potential energy difference of the test charge is given by Ue = Fe rf - Fe r0

Electric Potential Energy Difference

(Point Charge)

ELECTRIC POTENTIAL ENERGY DIFFERENCE - POINT CHARGE

=kq0Q

rf2

•rf -kq0Q

r02

•r0

Ue =kq0Q

rf

-kq0Q

r0

We define the electric potential difference V to be the potential energy per unit charge. Thus


ELECTRIC POTENTIAL DIFFERENCE - POINT CHARGE

Ue

q0

V = = kQ

rf

- kQ

r0

Electric Potential Difference

(Point Charge)

We can also define the gravitational potential difference Vg to be the potential energy per unit mass. Thus

Gravitational Potential Difference

(Point Mass)

UG

mVG = = GM

rf

- GM

r0

Note the missing word "Energy"



We eliminate the need for the presence of a test charge in the expression for electric potential energy.We essentially have a quantity that tells us the potential energy per unit positive charge.

Of course, at the surface of the earth we have a local gravitational field, represented with a bunch of g's:


Consider a baseball in a gravitational field:

If we raise the ball from a height of h0 to a height of hf and release it, we know it will relinquish the potential energy we gave to it - converting it to kinetic energy.The potential energy of the ball in the local gravitational field is given by

UG = FG hf - FG h0

= mghf - mgh0

h0

hf

h

Gravitational Potential Energy

Difference

GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - LOCAL

GRAVITATIONAL POTENTIAL DIFFERENCE - LOCALGravitational

Potential Difference

UG

mVG = = ghW

m=

A parallel plate capacitor is essentially two metal plates separated by a small distance.If we connect the two plates to a battery, electrons from the negative side of the battery will be "loaded" onto one plate, and electrons from the other plate will be drawn off of the other plate into the positive side of the battery.


ELECTRIC FIELD IN A PARALLEL PLATE CAPACITOR

+-

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

+

-

FYI: The E-field will be perpendicular to the plates. Why?

FYI: The E-field will act just like the local gravitational field.


Uq0

ELECTRIC POTENTIAL ENERGY DIFFERENCE - PLATES

V = = EdW

q0

=

The potential energy of a test charge in the local electric field is given by

UE = FE df - FE d0

= q0Edf - q0Ed0

Electric Potential Energy Difference

(Parallel Plates)

Electric Potential Difference

(Parallel Plates)

ELECTRIC POTENTIAL DIFFERENCE - PARALLEL PLATES


All right. Let’s do some practice problems.An electron in the vicinity of a proton is moved from a distance of 1 meter to a distance of 2 meters. What is the potential energy difference between initial and final positions?Note first that we seek the potential ENERGY difference.

Ue =kq0Q

rf

-kq0Q

r0

Ue = kq0Q1

rf

- 1

r0

Ue = (9×109)(-1.6×10-19)(1.6×10-19)12

- 11

Ue = +1.152×10-28 J

FYI: Note that the electron-proton system gained energy.

FYI: If the electron had begun at 2 m and ended at 1 m, our change in potential energy would have been negative. Why?

(use the point charge form)

FYI: Perhaps you recall that only CONSERVATIVE forces can have associated potential energies. The electric force, like the gravitational force is, in fact, conservative. This means that the potential energy difference does NOT depend on the path of the charge.


An electron in the vicinity of a proton is moved from a distance of 1 meter to a distance of 2 meters. What is the potential difference between initial and final positions?Potential difference is just potential ENERGY difference divided by q0

– in this case the charge of the electron:

Ue

q0

V = +1.152×10-28 J-1.6×10-19 C

= = -7.2×10-10 V

Note that the SI unit for electric potential difference is the joule / coulomb or volt.

FYI: We could have bypassed the previous problem and found V directly by using

Ue

q0

V = = kQ

rf

- kQ

r0

FYI: These are the same volts that you are familiar with.


An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (a) Find the potential energy lost by the electron as it travels to the positive plate.

Again, choose the correct formula – that for ENERGY:

UE = q0Edf - q0Ed0 (use the parallel plate form)UE = q0E(df - d0)

UE = (-1.6×10-19)(250)(0.02 - 0)

UE = -8×10-19 J


An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (b) Find the speed of the electron as it strikes the positive plate.

Use energy considerations.

K + UE = 0

K = -UE

K = 8×10-19 J12mv2 = 8×10-19

12(9.11×10-31)v2 = 8×10-19

v = 1.3×106 m/s

FYI: Don’t confuse voltage V with velocity v.


An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (c) Find potential difference between the two plates.

Pick the right formula:

V = Ed (use the parallel plate form)

V = (250)(0.02)V = 5 V


Find the total electrostatic potential energy in the charge configuration shown. Assume the charges have been assembled from infinity!

q1 = +5C q2 = +1C

q3 = -2C

0.25 m

0.25 m

0.25 m

Pick the right formula and work in pairs:

Ue =kqaqb

rf

-kqaqb

(use the point charge form)

0

U12 =kq1q2

r12

U13 =kq1q3

r13

U23 =kq2q3

r23

U12 =k510.25

U12 = 0.18 J

U13 =-k520.25

U13 = -0.36 J

U23 =-k120.25

U23 = -0.072 J

Utotal = 0.18 J + -0.36 J + -0.072 J = -0.252 J

FYI: The NEGATIVE sign means that these charges have a lower potential energy than they would if they were farther away. The highest potential energy for these particular charges would be ZERO, at .

FYI: If all the charges were POSITIVE (or NEGATIVE), the total potential energy would be positive. The lowest potential energy would occur at , and would be ZERO.


Two electrons are placed 1 cm apart and released. At what speed are they each traveling after their separation is 10 cm? Use energy considerations:

K + UE = 0

Kf + Uf = K0 + U0 0

12mv2 +

12mv2 +

kqq

rf

=kqq

r0

mv2 = kq2 1

r0

- 1

rf

v2 =kq2

m1

r0

- 1

rf

v2 =(9×109)(-1.6×10-19)2

9.11×10-31

1.01

- 1.10

v = 150.87 m/s

Question: How would you find their speeds after they have separated to infinity? Find their speed now. Is it what you expected?

topic 6.2 extended d – electric potential energy and p.d

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