Electric Fields Phys102 General Physics II Electric Fields

ƒ Topics – Electric field = Force per unit Charge – Electric Field Lines and Electric Flux – Electric field from more than 1 charge – Electric Dipoles – Motion of point charges in an electric field – Examples of finding electric fields from continuous charges

The Electric Field ƒ Definition of the electric field. Whenever charges are present and if I bring up another charge, it will feel a net Coulomb force from all the others. It is convenient to say that there is field there equal to the force per unit positive charge. E=F/q0. The direction of the electric field is along r and points in the direction a positive test charge would move. This idea was proposed by Michael Faraday in the 1830’s. The idea of the field replaces the charges as defining the situation. Consider two point charges: r

q1

r + q0

q1

The Coulomb force is F= kq1q0/r2 The force per unit charge is E = F/q0 and then the electric field at r is E = kq1/r2 due to the point charge q1 . The units are Newton/Coulomb. The electric field has direction and is a vector. How do we find the direction.? The direction is the direction a unit positive test charge would move. r

q0

E

If q1 were positive

q1

1

Point negative charge

q1

Electric Field Lines Like charges (++) Opposite charges (+ -)

r

E= kq1/r2

q1

This is called an electric dipole.

Electric Field Lines: a graphic concept used to draw pictures as an aid to develop intuition about its behavior.

Typical Electric Fields (SI Units) ƒ 1 cm away from 1 nC of negative charge – E = kq /r2 = 1010 *10-9/ 10-4 =105 N /C – N.m2/C2 C / m2 = N/C

r q

.E

Here are the drawing rules. ƒ E-field lines begin on + charges and end on - charges. (or infinity). ƒ They enter or leave charge symmetrically. ƒ The number of lines entering or leaving a charge is proportional to the charge ƒ The density of lines indicates the strength of E at that point. ƒ At large distances from a system of charges, the lines become isotropic and radial as from a single point charge equal to the net charge of the system. ƒ No two field lines can cross.

ƒ Fair weather atmospheric electricity = 100 N/C downward 100 km high in the ionosphere +++++ •E --------Earth ƒ Field due to a proton at the location of the electron in the H atom. The radius of the electron orbit is 0.5*10-10 m. – E = kq /r2 = 1010 *1.6*10-19/ (0.5 *10-10 )2 = 4*1011 N /C

Hydrogen atom

+

r

-

1N / C = Volt/meter

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Example of field lines for a uniform distribution of positive charge on one side of a very large nonconducting sheet. This is called a uniform electric field. How would the electric field change if both sides were charged? How would things change if the sheet were conducting?

Methods of evaluating electric fields ƒ Direct evaluation from Coulombs Law or brute force method If we know where the charges are, we can find E from E = ∑ kqi/ri2. This is a vector equation and can be complex and messy to evaluate and we may have to resort to a computer. The principle of superposition guarantees the result. ƒ Instead of summing the charge we can imagine a continuous distribution and integrate it. This distribution may be over a volume, a surface or just a line. – E = ∫dE = ∫ kdqi/r2 r where r is a unit vector directed from charge dq to the field point. – dq = ρdV , or dq =σ dA, or dq = λ dl

Electric field due to a point charge ƒ Direct evaluation from Coulombs Law or brute force method If we know where the charges are, we can find E from E = ∑ kqi/ri2. This is a vector equation and can be complex

F 1 |q| E= = q0 4πε 0 r 2

Superposition principle

E=

F 1 |q| = q0 4πε 0 r 2

G G G G G F = F1 + F2 + F3 + ....... + Fn

G G G G G G F F1 F2 F3 Fn E= = + + + ....... + q0 q0 q0 q0 q0 G G G G = E1 + E2 + E3 + ......... + En

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Example of finding electric field from two charges We have q =+10 nC at the origin, q = +15 nC at x=4 m. 1 2 What is E at y=3 m and x=0? point P y

E

Field due to q1

5

3

E = 1010 N.m2/C2 10 X10-9 C/(3m)2 = 11 N/C in the y direction.

φ 4

q1=10 nc

Field due to q2

x

4

q1=10 nc

x q2 =15 nc

Now add all components

E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C

q2 =15 nc

at some angle φ Resolve into x and y components

Use principle of superposition Find x and y components of electric field due to both charges and add them up

E

Example continued Ey= 11 + 3.6 = 14.6 N/C

Ey= 11 + 3.6 = 14.6 N/C Ex= -4.8 N/C

Ey=E sin φ = 6 ∗ 3/5 =18/5 = 3.6 Ν/C

Magnitude

Ex=E cos φ = 6 ∗ (−4)/5 =−24/5 = −4.8 Ν/C

Example of two identical charges on the x axis. What is the field on the y axis?

E = Ex2 + E y2

Example of two opposite charges on the x axis. What is the filed on the y axis? y

y E y

φ1

Εx

3

q1=10 nc

4

x

2

4

Using unit vector notation we can also write the electric field vector as:

E = Ex2 + E y2

(14.6 ) + ( −4.8)

3θ

q2 =15 nc

Magnitude of electric field

2

φ

Ex= 0 3

Ex= -4.8 N/C

y

Recall E =kq/r2 and k=8.99 x 109 N.m2/C2

Ey= 11 N/C

P

E=

Example continued

= 15.4 N / C

∧ ∧ G E = −4.8 i + 14.6 j

q2 =15 nc

5 4

3θ

φ

x

4 q2 = -15 nc

q2 =15 nc

5 4

φ

x

q2 =15 nc

E = 1010 N.m2/C2 15 X10-9 C/(5m)2 = 6 N/C

Ey=2*E sin φ = 2∗6 ∗ 3/5 = 7.2 N/C Εx=0

Εy=0 Ex=2*E cos φ = 2∗6 ∗ 4/5 = - 9.6 N/C

φ1 = atan Ey/Ex= atan (14.6/-4.8)= 72.8 deg

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4 equal charges symmetrically spaced along a line. What is the field at point P? (y and x = 0)

Electric field due to an Electric Dipole

y

ƒ Direct evaluation from Coulombs Law or brute force method If we know where the charges are, we can find E from

P r1

r2 θ2

θ1 q1

θ4 θ3 r 3

r4

φ q3

q2

E=

x q4 ∞

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E y = k ∑ ∆qi cos θ i / ri 2

E y = k ∑ qi cos θ i / ri 2 i =1

i =1

Find electric field due to a line of uniform + charge of length L with linear charge density equal to λ y

dE

dE = k dq /r2 dEy

dEx -x -L/2

x

dq

E y = kλ ∫− L / 2cosθ dx /r 2 L /2

θ0

∫ θ cos θ dθ −

0

L/2

Ex =

dEy= dE cos θ +x L/2

dEy= k λ dx cos θ /r2

Ey = kλ / y

What is the electric field from an infinitely long wire with linear charge density of +100 nC/m at a point 10 away from it. What do the lines of flux look like?

∫

dEx = 0

+++++++++++++++++++++++++++++++++++++++++

−L/2

r

y θ 0

F 1 |q| = q0 4πε 0 r 2

dq = λdx

Ey = Ey= k λ q cos θ /r2 for a point charge

dx/r2 = dθ/y

Ey =

2kλ sin θ 0 y

x= y tanθ r =y sec θ

sin θ 0 =

dx = y sec2 θ dθ r2 =y2sec2 θ

L /2

Ey =

.

2kλ sinθ 0 y 10

y =10 cm Ey

2

−9

2 *10 Nm *100 *10 C / m sin90 = 2 *10 4 N /C 0.1m

Typical field for the electrostatic smoke cleaner

y 2 + L2 /4

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Back to computing Electric Fields ƒ

Field due to arc of charge L/2

Electric field due to a line of uniform charge

Ey =

∫

dE y = 0

−L / 2

ƒ Electric field due to an arc of a circle of uniform charge.

dEx= k dq cos θ /r2 dEx= k λ ds cos θ /r2

ƒ Electric field due to a ring of uniform charge

Ex = k λ

ƒ Electric field of a uniform charged disk

L/2

∫

rdθ cos θ / r = k λ / r 2

Ex =

∫ dθ cos θ

s=r θ ds=r dθ

−θ0

−L/2

ƒ Next we will go on to another simpler method to calculate electric fields that works for highly symmetric situations using Gauss’s Law.

θ0

2kλ sinθ 0 r

What is the field at the center of a circle of charge? Ans. 0

Find the electric field on the axis of a uniformly charged ring with linear charge density λ = Q/2π Q/2πR.

Ez = ∫ dE cos θ Ez = Ez = dq = λds

k λ cos θ ∫ ds r2 k λ cos θ 2π R r2

kQz Ez = 2 ( z + R 2 )3/ 2

dE = k

Electric field due to a charged disk

dq λ ds =k 2 r2 r

2π

2π

0

0

∫ ds = ∫ Rdθ = R ∫ dθ = 2π R s = Rθ r2 =z2+R2 cos θ =z/r =0 at z=0 =0 at z=infinity =max at z=0.7R

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Measuring the elementary charge

Electric field gradient

ƒ When a dipole is an electric field that varies with position, then the magnitude of the electric force will be different for the two charges. The dipole can be permanent like NaCl or water or induced as seen in the hanging pith ball. Induced dipoles are always attracted to the region of higher field. Explains why wood is attracted to the teflon rod and how a smoke remover or microwave oven works.

Electrostatic smoke precipitator model ƒ Negatively charged central wire has electric field that varies as as 1/r (strong electric field gradient). Field induces a dipole moment moment on the smoke particles. The positive end gets attracted more to the wire ƒ In the meantime a corona discharge is created. This just means that induced dipole moments in the air molecules cause them to be attracted towards the wire where they receive an electron and get repelled producing a cloud of ions around the wire. ƒ When the smoke particle hits the wire it receives an electron and then is repelled to the side of the can where it sticks. However, it only has to enter the cloud of ions before it is repelled. ƒ It would also work if the polarity of the wire is reversed

Electric Dipoles: A pair of equal and opposite charges q separated by a displacement d is called an electric dipole. It has an electric dipole moment p=qd p=qd.. IEI ~ 2kqd/r3 when r is large compared to d p=qd = the electric dipole moment •P

+ +q d

p - -q

r IEI ~2kp/r3 Note inverse cube law IEI = kq/r2 Note inverse square law for a single charge.

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Water (H2O) is a molecule that has a permanent dipole moment. GIven p = 6.2 x 10 - 30 C m

H2O in a Uniform Electric Field

And q = -10 e and q = +10e

What is d? d = p / 10e = 6.2 x 10 -30 C m / 10*1.6 x 10 -19 C = 3.9 x 10 -12 m Very small distance but still is responsible for the conductivity of water.

There exist a torque on the water molecule To rotate it so that p lines up with E. x

Torque about the com = τ F x sin θ + F(d-x)sin θ = Fdsin θ = qEdsin θ = pEsin θ = p x E

When a dipole is an electric field, the dipole moment wants to rotate to line up with the electric field. It experiences a torque.

τ=pxE Potential Energy U = -W = -pEcosθ = − p. Ε Is a minimum when p aligns with E

Leads to how microwave ovens heat up food

Motion of point charges in electric fields ƒ When a point charge such as an electron is placed in an electric field E, it is accelerated according to Newton’s Law: a = F/m = qE/m for uniform electric fields

Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s. How much is the electron deflected after traveling 1 cm. V •e d

a = F/m = mg/m = g for uniform gravitational fields If the field is uniform, we now have a projectile motion problemproblemconstant acceleration in one direction. So we have parabolic motion just as in hitting a baseball, etc except the magnitudes of velocities and accelerations are different. Replace g by qE/m in all equations; For example, In y =1/2at2 we get y =1/2(qE/m)t2

E

E

Since velocity in x direction does not change, t=d/v =10-2/106 = 10-6 sec, so the distance the electron falls upward is y =1/2at2 = 0.5*eE/m*t2 = 0.5*1.6*10-19*2*103/10 - 30*(10-8)2 = 0.016m

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WarmWarm-up set 2

Kelvin Water Drop Generator Am. J. Phys. 68,1084(2000)

1. [153709] Can there be an electric field at a point where there is no charge? Can a charge experience a force due to its own field? Please write a one or two sentence answer for each question. 2. [153707] An insulator is a material that... three are correct is not penetrated by electric fields none of these cannot carry an electric charge cannot feel an electrical force 3. [153708] Which of the following is true of a perfect conductor ? There can be no electric charge on the surface. There cannot be an electric field inside. There cannot be any excess electric charge inside. There cannot be any electric charges inside. Two of the choices are correct

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