3/26/2015 1

University Physics (PHYS 2326)

Lecture 4

Electrostatics

Electric flux and Gauss’s law

Electrical energy

potential difference and

electric potential

potential energy of charged

conductors

Chapters 22-23

http://www2.swgc.mun.ca/physics/physlets/jepoten2.html

The Coulomb force is a conservative force

A potential energy function can be defined for any conservative force, including Coulomb force

The notions of potential and potential energy are important for practical problem solving

3/26/2015 2

3/26/2015 3

The electrostatic force is conservative

As in mechanics, work is

Work done on the positive charge by moving it from A to B

3/26/2015 4

A B

E

d

cosW Fd

cosW Fd qEd

The work done by a conservative force equals the negative of the change in potential energy, DPE

This equation is valid only for the case of a uniform electric field

allows to introduce the concept of electric potential

3/26/2015 5

PE W qEdD

The potential difference between points A and B, VB-VA, is defined as the change in potential energy (final minus initial value) of a charge, q, moved from A to B, divided by the charge

Electric potential is a scalar quantity Electric potential difference is a measure of electric

energy per unit charge Potential is often referred to as “voltage”

3/26/2015 6

B A

PEV V V

q

DD

Electric potential difference is the work done to move a charge from a point A to a point B divided by the magnitude of the charge. Thus the SI units of electric potential

In other words, 1 J of work is required to move a 1 C of charge between two points that are at potential difference of 1 V

3/26/2015 7

1 1V J C

Units of electric field (N/C) can be expressed in terms of the units of potential (as volts per meter)

Because the positive tends to move in the direction of the electric field, work must be done on the charge to move it in the direction, opposite the field. Thus, A positive charge gains electric potential energy when it is

moved in a direction opposite the electric field

A negative charge looses electrical potential energy when it moves in the direction opposite the electric field

3/26/2015 8

1 1N C V m

The same kinetic-potential energy theorem works here

If a positive charge is released from A, it accelerates in the direction of electric field, i.e. gains kinetic energy

If a negative charge is released from A, it accelerates in the direction opposite the electric field

3/26/2015 9

A

B

q d

A

B

m d

E g

i i f fKE PE KE PE

3/26/2015 10

Vab

What is the speed of an electron accelerated from rest across a

potential difference of 100V? What is the speed of a proton

accelerated under the same conditions?

Given:

DV=100 V

me = 9.1110-31 kg

mp = 1.6710-27 kg

|e| = 1.6010-19 C

Find:

ve=?

vp=?

Observations:

1. given potential energy

difference, one can find the

kinetic energy difference

2. kinetic energy is related to

speed

i i f fKE PE KE PE

f i fKE KE KE PE q V D D

21 2

2f f

q Vmv q V v

m

D D

6 55.9 10 , 1.3 10e pm mv v

s s

Electric circuits: point of zero potential is defined by grounding some point in the circuit

Electric potential due to a point charge at a point in space: point of zero potential is taken at an infinite distance from the charge

With this choice, a potential can be found as

Note: the potential depends only on charge of an object, q, and a distance from this object to a point in space, r.

3/26/2015 11

e

qV k

r

If more than one point charge is present, their electric potential can be found by applying superposition principle

The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges.

Remember that potentials are scalar quantities!

3/26/2015 12

Consider a system of two particles

If V1 is the electric potential due to charge q1 at a point P, then work required to bring the charge q2 from infinity to P without acceleration is q2V1. If a distance between P and q1 is r, then by definition

Potential energy is positive if charges are of the same sign and vice versa.

3/26/2015 13

P A

q1 q2

r

1 22 1 e

q qPE q V k

r

3/26/2015 14

Three ions, Na+, Na+, and Cl-, located such, that they

form corners of an equilateral triangle of side 2 nm in

water. What is the electric potential energy of one of the

Na+ ions?

Cl-

Na+ Na+

? Na Cl Na Na Na

e e e Cl Na

q q q q qPE k k k q q

r r r

but : !Cl Naq q

0Nae Na Na

qPE k q q

r

Recall that work is opposite of the change in potential energy,

No work is required to move a charge between two points that are at the same potential. That is, W=0 if VB=VA

Recall: 1. all charge of the charged conductor is located on its surface 2. electric field, E, is always perpendicular to its surface, i.e. no

work is done if charges are moved along the surface

Thus: potential is constant everywhere on the surface of a charged conductor in equilibrium

3/26/2015 15

B AW PE q V V

… but that’s not all!

Because the electric field in zero inside the conductor, no work is required to move charges between any two points, i.e.

If work is zero, any two points inside the conductor have the same potential, i.e. potential is constant everywhere inside a conductor

Finally, since one of the points can be arbitrarily close to the surface of the conductor, the electric potential is constant everywhere inside a conductor and equal to its value at the surface!

Note that the potential inside a conductor is not necessarily zero, even though the interior electric field is always zero!

3/26/2015 16

0B AW q V V

A unit of energy commonly used in atomic, nuclear and particle physics is electron volt (eV)

The electron volt is defined as the energy that electron (or proton) gains when accelerating through a potential difference of 1 V

Relation to SI:

1 eV = 1.6010-19 C·V = 1.6010-19 J

3/26/2015 17

Vab=1 V

Remember that potential is a scalar quantity Superposition principle is an algebraic sum of potentials due

to a system of charges

Signs are important

Just in mechanics, only changes in electric potential are significant, hence, the point you choose for zero electric potential is arbitrary.

3/26/2015 18

3/26/2015 19

In the Bohr model of a hydrogen atom, the electron, if it is in the

ground state, orbits the proton at a distance of r = 5.2910-11 m. Find

the ionization energy of the atom, i.e. the energy required to remove

the electron from the atom.

Note that the Bohr model, the idea of electrons as tiny balls orbiting the nucleus, is not

a very good model of the atom. A better picture is one in which the electron is spread

out around the nucleus in a cloud of varying density; however, the Bohr model does

give the right answer for the ionization energy

3/26/2015 20

In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,

orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,

i.e. the energy required to remove the electron from the atom.

Given:

r = 5.292 x 10-11 m

me = 9.1110-31 kg

mp = 1.6710-27 kg

|e| = 1.6010-19 C

Find:

E=?

The ionization energy equals to the total energy of the

electron-proton system,

E PE KE

22 2182.18 10 J -13.6 eV

2 2

ee e

k ee m eE k k

r mr r

The velocity of e can be found by analyzing the force

on the electron. This force is the Coulomb force;

because the electron travels in a circular orbit, the

acceleration will be the centripetal acceleration:

c cma F

2 2

,2

e

e vPE k KE m

r with

or

2 2

2,e

v em k

r r or

22 ,e

ev k

mr

Thus, total energy is

They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage)

The electric field at every point of an equipotential surface is perpendicular to the surface

3/26/2015 21

convenient to represent by drawing

equipotential lines

3/26/2015 22

3/26/2015 23

ˆˆ ˆ( )dV V V V

E i j kdr x y z

3/26/2015 24

23.11 .. Three point charges, which initially are infinitely far apart, are placed at the

corners of an equilateral triangle with sides d.Two of the point charges are identical and

have charge q. If zero net work is required to place the three charges at the corners of

the triangle, what must the value of the third charge be?

23.27 .. A thin spherical shell with radius R1 = 3.00cm is concentric with a larger

thin spherical shell with radius R2 = 5.00cm. Both shells are made of insulating

material. The smaller shell has charge q1 = +6.00nC distributed uniformly over its

surface, and the larger shell has charge q2 = -9.00nC distributed uniformly over its

surface. Take the electric potential to be zero at an infinite distance from both shells.

(a) What is the electric potential due to the two shells at the following distance from

their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00cm? (b) What is the

magnitude of the potential difference between the surfaces of the two shells? Which

shell is at higher potential: the inner shell or the outer shell?

3/26/2015 25

1. a) E = 0 V/m in throughout some region of space, can you conclude that

the potential V = 0 in this region?

b) V = 0 V throughout some region of space. Can you conclude that the

electric field E = 0 V/m in this region?

--Find the electric potential everywhere for a sphere (radius

R) with charge (Q) uniformly distributed. Take V=0 at infinity.

--Sketch V vs r and Er vs r.

 

E =

KQ

R3r ˆ r

KQ

r2ˆ r

r < R

R < r

ì

í ï

î ï

Given

3/26/2015 26

Find the electric potential everywhere for a sphere (radius R) with charge (Q) uniformly

distributed.

 

V =

3KQ

2R-

KQ

2R3r2

KQ

r

(r < R)

(R < r)

ì

í

ï ï

î

ï ï

 

E r =K

Q

R3r

KQ

r2

(r < R)

(R < r)

ì

í ï

î ï

3/26/2015 27

Find the x,y and z components of the electric field, given that the electric potential of a

disk is given by

 

Vdisk =Q

2pR2e0

z2 + R2 - z( )

3/26/2015 28

Find the z component of the electric field, given that the electric potential of a disk is

given by

 

Vdisk =Q

2pR2e0

z2 + R2 - z( )

Edisk( )z

= -dV

dz=

h

2e0

1-z

z2 + R2

æ

èç

ö

ø÷

3/26/2015 29

is perp to equipotential surfaces

points downhill (decreasing V)

--strength proportional to spacing equipotentials

Geometry of potential/field

3/26/2015 30

--field is zero inside conductor

--field is perpendicular at surface

--conductor is at equipotential (no work to move)

Conductor in equilibrium: field and potential

3/26/2015 31

--equipotentials are parallel

to nearby conductor

Conductor in equilibrium: equipotentials

3/26/2015 32

--Find the electric potential everywhere for a point charge (q) at the center of a hollow

metal sphere (inner radius a, outer radius b) with charge Q. (Take V = 0 at infinity.)

--Sketch V vs r and Er vs r.

a

b

Problem: Finding Potential

 

E r(r) =

Kq

r2, r < a

0, a < r < b

K(Q + q)

r2, b < r

ì

í

ï ï

î

ï ï

3/26/2015 33

Problem: Finding Potential (ans)

 

E r(r) =

Kq

r2, r < a

0, a < r < b

K(Q + q)

r2, b < r

ì

í

ï ï

î

ï ï

r

 

d

r Vr V∞

∆V

Vr

Va

∆V

 

E

For all r

For b < r

 

= - -K(Q + q)

r

é

ë ê ù

û ú ¥

r

 

=K(Q + q)

r-

K(Q + q)

¥

 

Defined V¥ = 0

 

Vr - V¥ =K(Q + q)

r

 

Vr =K(Q + q)

r(R < r)

Finding Vr

For a < r < b

 

from b < r; Vb =K(Q + q)

b

 

Vr =K(Q + q)

b(a < r < b)

For r < a

 

= - -Kq

r

é

ë ê ù

û ú a

r

 

=Kq

r-

Kq

a

 

From before Va = Vb =K(Q + q)

b

 

Vr -K(Q + q)

b=

Kq

r-

Kq

a

 

Vr =K(Q + q)

b+

Kq

r-

Kq

a

æ

è

ö

ø (r < a)

3/26/2015 34

 

V =

Kq + Q

r

Kq + Q

b

Kq + Q

b+ Kq

1

r-

1

a

æ

è

ö

ø

ì

í

ï ï ï

î

ï ï ï

b < r

a < r < b

r < a

not origin

Problem: finding

Potential (Answer)

 

E r(r) =

Kq

r2, r < a

0, a < r < b

K(Q + q)

r2, b < r

ì

í

ï ï

î

ï ï

3/26/2015 35

-charge separation

-not sustained

Sources of potential: Capacitor

3/26/2015 36

definition

 

E =

KQ

R3r ˆ r

KQ

r2ˆ r

r < R

R < r

ì

í ï

î ï

For all r

For R < r

 

= - -KQ

r

é

ë ê ù

û ú ¥

r

 

=KQ

r-

KQ

¥

 

Defined V¥ = 0

 

Vr - V¥ =KQ

r

 

Vr =KQ

r(R < r)

Finding Vr

For r < R

 

= -KQ

2R3r2é

ë ê ù

û ú R

r

 

= -KQ

2R3r2 -

KQ

2R3R2æ

è

ö

ø

 

From R < r : VR =KQ

R=

KQ

R3R2

 

Vr -KQ

R= -

KQ

2R3r2 -

KQ

2R3R2æ

è

ö

ø

 

Vr =3KQ

2R3R2 -

KQ

2R3r2 (r < R)

r

∆V

Vr

Vr

 

E

 

d

r Vr V∞

VR

∆V