Electrostatics

Point Charges

• As we did for masses in mechanics, we consider charges to be single points in space

• Charges can be single protons, electrons, or ions, as well as larger clumps of them, or even larger surfaces that have a net positive or negative charge

• Charges are measured in Coulombs (unit denoted by C). 1 C=6.242×1018 protons

Interactions Between Point Charges

• Recall the gravitational force between two bodies is expressed as

• Likewise the electrostatic force between two point charges is the analogous

• This relation is referred to as Coulomb’s Law• The point masses in the Law of Universal

Gravitation above are analogous to the point charges in Coulomb’s law

vF =−

GmMr2 r

vF =

kq1q2

r 2 r

But what is k?

• Lets look at Coulomb’s Law again• Since both q’s are measured in Coulomb’s,

denoted by C, and the units of Force are in Newtons, the unit analysis yields

• Solving for k we find that k’s units are in

• Infact

vF =

kq1q2

r 2 r

N =kC2

m2

k =Nm2

C2

k =9×109 N⋅m2

C2

Electric Fields

• Near the end of his life British experimental physicist Michael Faraday suggested the idea that electric forces extend into all regions of space around a conductor or point charge

• This idea was rejected by most of his contemporaries, but well accepted by the late 19th century. The work of James Clerk Maxwell, mathematically finalized these ideas.

So what are Electric Fields?

• The notion is any point charge in space induces a field into the surrounding area.

• A field assigns a vector to every point in space• The Electric field on its own does nothing to

affect surrounding matter unless there is a net charge on said objects.

• If there is a net charge on surrounding objects,then

rF =Q

rE =

kQqr2 r

What does it look like?

Electric Field

• Electric Fields are radial, and are a function of the distance away from the charge, just like Coulomb’s Law

• Electric Fields are a property of space, and don’t imply that any forces are present in a system only the potential for Forces

rE =

kQr2 r and

rF =Q

rE

Superposition

• Electric fields in space are the additive sum of all electric fields caused by point charges

• In other words, with vector addition

• Also when two point charges exist in space near each other they exert a force equal to

rEi∑ =

rEtotal

rF =Q

rE or

rF =

kQqr2 r

What does this look like?

Field Factualism

• The force immediately between the two positive charges interacting for any sign charge is 0.

• The field at any point in space is the sum of all fields generated by however many point charges exist in local space

• Electric fields are conservative, that is , like the gravitational field they obey

U f −Ui =−W =−

rFgd

rr∫

Fundamental Theorem

• Conservative Force fields possess a unique potential energy function U(r) such that

• Remember when change in potential energy is positive, work is negative and visa versa

• Lets examine some examples

U f −Ui =U(b)−U(a) =−

rFgd

rs

a

b

∫

The integral and you

• In order to do anything in electromagnetism we need to be able to integrate, or at least know

• Remember with a conservative field the special function U(r) determines potential energy, independent of path– if you know it you will go far!

−

rFgd

rr∫ = −

kQq

r2a

b

∫ dr =kQq

r|ab=kQq

b−kQq

a=U(b) −U(a)

Potential Changes

• Remember total energy is constant• Two positive charges moving away from each

other like so

Experience a potential change of

+q +Qr =2m

+q+Q

r =10m

U f −Ui =kQq10

−kQq2

=kQq10

−5kQq10

=−4kQq10

=−2kQq5

Signs

• This negative change in potential indicates a decrease in potential energy as the negative charges move away form each other and an increase in kinetic energy

• Think of a ball rolling down a hill– it becomes faster the farther it falls in height, but it has less potential to continue rolling, as it is approaching the bottom of the hill

Potential Changes

• Remember total energy is constant• One positive and one negative charge moving

away from each other like so

Experience a potential change of

−q +Qr =2m

−q+Q

r =10m

U f −Ui =kQ(−q)10

−kQ(−q)

2=−

kQq10

+5kQq10

=4kQq10

=2kQq5

Signs

• This positive change in potential energy is because we are forcing apart positive and negative charges– the system stores energy much like a spring

• As the charges are allowed to come together due to their attraction, potential energy is changed to kinetic energy

Potential Energy and Work

• Two positive charges coming together

• Two positive charges moving apart

• One positive and one negative charge moving apart

• One positive and one negative charge moving together

W < 0 and ΔU > 0

W > 0 and ΔU < 0

W < 0 and ΔU>0

W > 0 and ΔU < 0

The Potential Function

• Remember, all conservative fields have a “magic” function U(r) which induces path independence

• U(r) however is dependent on the point charge being analyzed

• We wish to generalize a function that is applicable to all charges

• This function exists at all points in space due to a point charge and is called the Potential Function

The Mathematics

• The mathematics are similar to what we have done before–

• The conclusion we have basically arrived at is

U f −Ui =−rF∫ gd

rr ; but remember

rF =Q

rE and then

U f −Ui =− QrEgd

rr =−Q

rEgd

rr∫∫

U f −Ui

Q=ΔUQ

=−rEgd

rr∫ =ΔV

ΔV = −

rEgd

rr∫ = −

kQ

r2∫ dr =kQ

r

Potent Potentials

• So the potential function, that exists at all points in space r distance from a point charge is

• Notice that for a negative charge we have negative potential values, for a positive charge we have positive ones

• By CONVENTION we dictate that• This is convention because we could have chosen

another integration constant which would have changed

V (r) =kQr

V (∞) =0

V (r) =kQr

+C

V(∞) =C

Breaking It downForce and Electric Field Potential Energy and Potential

Vector Quantity Scalar Quantity

Returns vector at all points in space Returns scalar at all points in space

E has units of N/C V has units of J/C, also called a volt.

Superposition applies Superposition applies

Electric field does nothing without the presence of another charge other than the one generating it

Potential does nothing without the presence of another charge other than the one generating it

In the presence of an Electric field, a Force is induced in charges present in the electric field

In the presence of potential a Potential Energy is induced in charges present in the potential field

rF =Q

rE U =QV

rEi∑ =

rEtotal Vi∑ =Vtotal

F =−dUdr

and E =−dVdr

ΔV = V (b) −V (a) = −

rEgd

rr

a

b

∫

Problem Solving Particulars

• Observe the following fixed charge arrangement

• In the middle

r =2mV = Vi∑ =kQr

+kQr

+kQr

+kQr

=4kQ

r=2kQ

So then for each charge

U =QV =2kQ2

+Q +Q

+Q +Q

Problem Solving Particulars

• This means that if one ofthe charges were releasedthey would be pushed to a point very far awayeventually reaching speeds of

r =2m

+Q +Q

+Q +Q

TE =PE + KE

TE =2kQ2

ΔU =U f −Ui =U(∞)−U(2) =−2kQ2

W =2kQ2 =12

mv2

v=4kQ2

m=2Q

km

Problem Solving Particulars

• Now lets look at the electric field• By superposition,

• Observe that the force vectorsoppose each other for Diagonal Q charges, so E=0In the middle.

r =2m

+Q +Q

+Q +Q

rEi∑ =

rEtotal

Problem Solving Particulars

• More mathematically speaking

r =2m

+Q +Q

+Q +Q

rE =

kQr2

rE1 =

kQ4

22

i −kQ4

22

j =kQ 28

i −kQ 28

j

rE2 =−

kQ4

22

i −kQ4

22

j =−kQ 28

i −kQ 28

j

rE3 =

kQ4

22

i +kQ4

22

j =kQ 28

i +kQ 28

j

rE4 =−

kQ4

22

i +kQ4

22

j =−kQ 28

i +kQ 28

jrE =0i + 0 j =0∑

What does it mean?

• This means that any charge -q placed at the middle of the square would experience 0 force but have a potential energy of

• Think of a ball sitting on top of a hillany push will cause it to fall

• It will lose potentialenergy and gain kinetic

U =−2kQq

What does it mean?

• This means that any charge q placed at the middle of the square would experience 0 force but have a potential energy of

• Think of a ball resting in a hole • A push will cause it to

oscillate in the holeunless its enoughto allow it to escape

U =2kQq

Class problem solving

• Draw the potential function in the middle of the two charges here versus the Y axis, and mathematically create the potential functionU(x,y) in terms of x and y. Also find E(y).

−Q −Qx

y(−a,0) (a,0)

Answers Aplenty

• So each charge contributes exactlyto the potential function

• Remember the• In this case each charge has a distance of

to the y axis at any point y.

Then by superposition

You can see that as y becomes larger the potential gets smaller. Therefore minimum potential occurs at V(y=0)

V =−kQ

rdistance =r = (x2 −x1)

2 + (y2 −y1)2

r = a2 + y2

Vi∑ =−2kQ

a2 + y2

Answers Aplenty

• This Potential function then looks a lot like a hole • The potential energy function however for a

particle with charge +Q or –Q however depends on the charge of Q. After all

• So for a positive Q, placed in the middle, the potential energy function is like a hole in the ground

• For a –Q charge, the potential energy function is like a hill

U =±QV

More Answers Aplenty

• We can easily find the electric field along the y axis with the potential function we have created– after all

• This only talks about the electric field as a function of y (along the y axis only)

• Notice E=0 at the origin since y =0. (the charges field contributions cancel each other out)

−dV

dr= E and V =

2kq

a2 + y2 simply take the derivative

rE =

4kqy

(a2 + y2 )3/2

More Answers Aplenty

• We can also easily determine where the electric field is pointing. Since fields point toward negative charges, in the +y values it points down, and in the –y values it points up

−Q −Qx

y(−a,0) (a,0)

More Classwork

• Explain what occurs if I were to release a negative charge of –q at the origin, with velocity of 2m/s in the +y direction. What speed in terms of a would it reach when it is very far away?

−Q −Qx

y(−a,0) (a,0)

More Answers

• A –q charge released at the origin with velocity 0 would not move– since we have velocity ofhowever, the negative charge will feel a net upward force, and

v =(2m/ s) j

ΔU = −qΔV = −q(V (∞) −V (0)) = −q 0 −−2kQ

a⎛⎝⎜

⎞⎠⎟

=2kQq

a

(remember that V =−2kQ

a2 + y2 so V (0) =

−2kQ

a)

• So we have that the total energy change as the negative charge moves from the origin to infinity is

• Since total energy is constant under only conservative forces then

ΔU =2kQq

a

Constant =KEi + PEi =KE f + PE f

Constant=12

mvi2 +

2kQqa

=12

mv2f + 0

(remember PE f =PE(∞) =U(∞) =0

then solving for vf ,

mvi2 +

4kQqa

=mvf2

vi2

m+4kQqma

=vf =4m

+4kQqma

=4a+ 4kQq

mawhere m is the mass of the charge in question

Yet More Classwork

• For the fixed charge arrangement below take the middle of the arrangement to be the origin

• Find the following-The electric field at the origin-The potential at the origin-The behavior of a +1C chargePlaced at the origin-The acceleration of such a charge at theOrigin.

+.1C +2.1C

−1.1C +.1C

E1 E2

E3 E4

r =0.2m

Vector Sums

• First find the E field from each charge individually. Recall that

• The angles in a square are all 45 degrees so

Ex =Ecosθ and Ey =Esinθ

sinθ =cosθ =22

rE1 =

kQr2 cosθ,−

kQr2 sinθ⎛

⎝⎜⎞⎠⎟=

kQ 22r2 ,−

kQ 22r2

⎛

⎝⎜⎞

⎠⎟= 1.6 ×1010 ,−1.6 ×1010( )

NC

rE2 = −

kQr2 cosθ,−

kQr2 sinθ⎛

⎝⎜⎞⎠⎟= −

kQ 22r2 ,−

kQ 22r2

⎛

⎝⎜⎞

⎠⎟= −3.34 ×1011,−3.34 ×1011( )

NC

rE3 =

kQr2 cosθ,

kQr2 sinθ⎛

⎝⎜⎞⎠⎟=

kQ 22r2 ,

kQ 22r2

⎛

⎝⎜⎞

⎠⎟= 1.75 ×1011,1.75 ×1011( )

NC

rE1 = −

kQr2 cosθ,

kQr2 sinθ⎛

⎝⎜⎞⎠⎟= −

kQ 22r2 ,

kQ 22r2

⎛

⎝⎜⎞

⎠⎟= −1.6 ×1010 ,1.6 ×1010( )

NC

rE∑ = −1.59 ×1011,−1.59 ×1011( )

NC

so the electric field points in the negative x and negative y direction

Potential at the Origin

• The potential function also obeys superposition so

• Remember this is NOT a vector. Electric field IS a vector.

Vi∑ =Vtotal

V (0,0) =kQ1

r+

kQ2

r+

kQ3

r+

kQ4

r=

kr

Qi∑ =(9.0 ×109 N

C)

.2m(1.2C)

=5.4 ×1010 JC

•Point Charge Behavior

• The force on a 1C point charge placed at the origin would be the following

• The magnitude of this force isor

• The acceleration is just a=F/m, for whatever mass we have

• The charge will simply collide with the -1.1 C charge

rF =Q

rE =(1C) −1.59 ×1011,−1.59 ×1011( )

NC

= −1.59 ×1011,−1.59 ×1011( )N

rF = Fx

2 + Fy2

2.25 ×1011N

AP Problems

• We will now work on some simple AP problems regarding the above material