Gravitational Potential Energy Lift a book by hand (F ext ) at - - PDF document

Briefly review the concepts of potential energy and work.

• Potential Energy = U = stored work in a system
• Work = energy put into or taken out of system by forces
• Work done by a (constant) force F :

θ cos | || | r F r F W v v v v ∆ = ∆ ⋅ =

∆r F

θ

Gravitational Potential Energy

Lift a book by hand (Fext) at constant velocity.

Fext mg initial final

h

Fext = mg Wext = Fext h = mgh Wgrav = -mgh Define ∆U = +Wext = -Wgrav= mgh Note that get to define U=0, typically at the ground.

U is for potential energy, do not confuse with “internal energy” in Thermo.

Gravitational Potential Energy (cont)

For conservative forces Mechanical Energy is conserved. Gravity is a conservative force. Coulomb force is also a conservative force. Friction is not a conservative force. If only conservative forces are acting, then ∆EMech=0.

= ∆ + ∆ U EKin U E E

Kin Mech

+ =

Electric Potential Energy

Charge in a constant field

• - - - - - - - - - - - - -

+

E FField=qE FExt=-qE

+

∆r

Initial position Final position

∆Uelec = change in U when moving +q from initial to final position.

field ext i f

W W U U U − = + = − = ∆

r E q U r F W U

field field

v v v v ∆ ⋅ − = ∆ → ∆ ⋅ − = − = ∆

General case

What if the E-field is not constant?

r E q U v v ∆ ⋅ − = ∆

∫

⋅ − = ∆

f i

r d E q U v v

Integral over the path from initial (i) position to final (f) position.

Electric Potential Energy

Since Coulomb forces are conservative, it means that the change in potential energy is path independent.

∫

⋅ − = ∆

f i

r d E q U v v

Electric Potential Energy

Positive charge in a constant field

Electric Potential Energy

Negative charge in a constant field

Observations

• If we need to exert a force to “push” or “pull” against the

field to move the particle to the new position, then U

• increases. In other words “we want to move the particle to

the new position” and “the field resists”.

• If we need to exert a force to “hold” the particle so the field

will not move the particle to the new position, then U

• decreases. In other words, “the field wants to move the

particle”, and “we resist”.

• Both can be summarized in the following statement:

“If the force exerted by the field opposes the motion of the particle, the field does negative work and U increases,

• therwise, U decreases”

Potential Energy between two point charges

r +Q1 +Q2

Imagine doing work to move the objects from infinitely far apart (initial) to the configuration drawn above (final).

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field ext i f

W W U U U − = + = − = ∆

∫

⋅ − = − = ∆

f i field i f

r d F U U U v v

∫

⋅ − = − = ∆

f i i f

r d E q U U U v v

* Note that force is not constant over the path!

r +Q1 Consider Q1 fixed and move Q2 from infinity to r.

Potential Energy between two point charges

r +Q1 +Q2

∫

⋅ − = − = ∆

f i i f

r d E q U U U v v

E-field generated by Q1. Moving Q2 through the field.

∫

∞

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − = ∆

r i f

dx x Q Q U U U

2 1 2 4

1 πε r Q Q U U U

i f 2 1

4 1 πε = − = ∆

Potential Energy between two point charges

We also need to define the zero point for potential energy. This is arbitrary, but the convention is U=0 when all charged

• bjects are infinitely far apart.

r Q Q U U U

i f 2 1

4 1 πε = − = ∆

r Q Q U f

2 1

4 1 πε =

Ui =U(∞)= 0 by our convention. Potential Energy between two electric charges.

Potential Energy between two point charges

• E. Potential energy vs. distance

• E. Potential Energy of a charge

distribution

∑

= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + =

i i i

r q q r q r q r q q U

3 3 2 2 1 1

4 ... 4 πε πε For a continuous charge distribution, replace the sum by an integral Potential energy associated to the field produced by charges qi

r1 r2 r3 q1 q2 q3 q0

• E. Potential Energy of a charge

distribution

∑

<

=

j i ij j i Total

r q q U 4 1 πε

For a continuous charge distribution, replace the sum by an integral To calculate the TOTAL potential energy we have to consider all the fields produced by all the charges qi

• n the other(n-1) charges qj

r1 r2 r3 q1 q2 q3 q0

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r U(r) r U(r)

r Q kQ r U

2 1

) ( =

Both cases below are for two point charges separated by a distance r. Which graph is correct for two negative charges? A) Left Plot, B) Right Plot

CPS Question

Just like a compressed spring stores potential energy.

Earlier we found that not only using forces, but also electric fields was very useful.

r r Q Q F ˆ 4 1

2 2 1 12

πε = v r r Q E ˆ 4 1

2 1 1

πε = v r r Q Q E q F ˆ 4 1

2 1 2 1 12

πε = = v v

r +Q1 +Q2

• Electric Field is the force per unit of charge due to the presence of Q1
• Electric field from Q1 is there even if Q2 is not there.

All of the above are vectors!

Electric Field

We find a similarly useful thing with electric potential energy.

r Q Q U

2 1 12

4 1 πε = r Q V

1 1

4 1 πε = r Q Q qV U

1 2 1 12

4 1 πε = =

r +Q1 +Q2

• Electric potential is the electric potential energy per unit of charge due

to the presence of Q1

• Electric potential from Q1 is there even if Q2 is not there.

All of the above are scalars!

Electric Potential

+Q1

Electric Field is a vector associated with a source charge Q1. Units are [Newtons/Coulomb]. Electric potential is a scalar associated with a source charge Q1. Units are [Joules/Coulomb] or [Volts].

r r Q E ˆ 4 1

2 1 1

πε = v r Q V

1 1

4 1 πε =

Electric Potential

V = Voltage = Electric Potential Units are [Volts] = [Joules/Coulomb] U = Electric Potential Energy Units are [Joules] Electric Field Units=[Volts]/[meter]=[Newton]/[Coulomb]

Electric Potential of a point charge

r q V 4 1 πε =

Analogy: Electrical pressure or electrical "height" Positive charges want to get away from higher voltage towards lower voltage. Just like a gas wants to move from high to low pressure.

Electric Potential of a charge distribution

∑

= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + =

i i i

r q q r q r q r q q U

3 3 2 2 1 1

4 ... 4 πε πε For a continuous charge distribution, replace the sum by an integral Potential energy associated to the field produced by the charges qi Electric potential due to charges qi

∑

=

i i i

r q V 4 1 πε

r1 r2 r3 q1 q2 q3 q0 “test charge”

∫

⋅ − = ∆ = ∆

f i

r d E q U V v v

Electric Potential from Electric Field

∫ ∫

⋅ − = ⋅ − = = ∆

f i f i Ext

r d E q r d F W U v v v r

. Elevation is a scalar. Contour lines show paths of constant elevation. Gravitational potential VG=gh If I stand at a certain elevation I have a gravitational potential energy [Joules] = mgh=mVG

Equipotential surfaces

Electric potential (Voltage) is a scalar. Contour lines show paths of constant Voltage (equipotentials). If a charge q is at a certain Voltage, it has an electrical potential energy [Joules] = qV

Equipotential surfaces

Equipotential surfaces and field lines

• Equi-potential surfaces and field lines are always mutually

perpendicular

• The Field is not necessarily constant on equipotential surfaces
• Larger density of equipotentials means larger variations of V,

and larger values of |E|

• E. Field and E. potential

What is Electric Field? Electric Field is a vector that is analogous to a steepness vector. Steepness cannot have one number at a given position, it depends which direction you look (vector).

∫

⋅ − = − = ∆

f i i f

r d E V V V v v

∫

∞

⋅ − = ∞ −

r

r d E V r V v v ) ( ) (

∫

∞

⋅ − = = ∆

r

r d E r V V v v ) (

Given the Electric field vector as a function of position, we can compute Voltage. If we integrate the “steepness” over a path, we find the change in elevation. This does not depend on our path taken.

• E. Field and E. potential

Vi =V(∞)= 0 by our convention.

dz dV E dy dV E dx dV E dz E dy E dx E dV

z y x z y x

− = − = − = ⇒ + + = − ; ;

∫ ∫

− = ⋅ − = ∆

b a b a

dV r d E V v v

The Electric field vector is the rate of change

• f Voltage in a given direction.

z z V y y V x x V E ˆ ˆ ˆ ∂ ∂ − ∂ ∂ − ∂ ∂ − = v

• E. Field and E. potential

V E ∇ − = r v

The Electric field vector “potential gradient” http://www.falstad.com/vector2de/

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CPS question

Two identical charge, +Q and +Q, are fixed in space. The electric potential (V) at the point X midway between the charges is: A) Zero B) Non-Zero +Q +Q

Point X

E=0 V=? x V

+Q +Q

x x V E ˆ ∂ ∂ − = v

CPS Question

Drawn are a set of equipotential lines. Consider the electric field at points A and B. Which of the following statements is true?

0V 10V 20V 30V Point A Point B

A)|EA| > |EB| B)|EA| < |EB| C)|EA| = |EB| D)Not enough information given. E) None of the above

32

CPS Question

Two charges, +Q and -Q, are fixed in space. The electric field at the point X midway between the charges is: A) Zero B) Non-Zero +Q

• Q

Point X

V=0 E=? x V

+Q

• Q

x x V E ˆ ∂ ∂ − = v

33

CPS Question

+Q +Q

• Q
• Q

Point P

The magnitude of the electric field at point P is: A) Zero B) Non-Zero The magnitude of the voltage at point P is: A) Zero B) Non-Zero All points on conductor must be at the same electrical potential.

∫

⋅ − = ∆

b a

r d E V v v

Imagine point a at Voltage Va. Since E=0 everywhere inside the conductor (no steepness), integral to point b is always 0. ∆Vab=0.

• E. Potential inside conductors