Chapter 19Electric Force and Electric Field

Interaction of Electric Charge

Charging an object• A glass rod is rubbed

with silk• Electrons are transferred

from the glass to the silk• Each electron adds a

negative charge to the silk

• An equal positive charge is left on the rod

Charge

Unit: C, Coulomb

+ −

Electric ChargeElectric charge is one of the fundamental attributes of the particles of which matter is made.

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qproton = +e

qelectron = −e

e = 1.6 ×10−19C

Electric Field

Like charges (++)

Electric dipole:Opposite charges (+−)

Vectors are arrows

i

j rv1 =2i + j

rv3 =−i + j

rv2 =1.5i −0.5 j

rv4 =−2i − j

What are these vectors?

i

j rv1 =2i

rv3 =−i −2 j

rv2 =i −1.5 j

rv4 =−3i + 2 j

Magnitude of a vector= Length of the arrow

4

3 vv =4i + 3 j

vv = 42 + 32 =5

If rv =ai + bj

⇒ vv = a2 +b2

What are the magnitudes?

i

j rv1 =2i

rv3 =−i −2 j

rv2 =i −1.5 j

rv4 =−3i + 2 j

Magnitudes (solution)

i

j rv1 =2i ⇒

rv1 =2

rv3 =−i −2 j

⇒rv3 = (−1)2 + (−2)2

= 12 + 22 = 5 =2.24

rv2 =i −1.5 j

⇒rv2 = 12 +1.52 =1.80

rv4 =−3i + 2 j

⇒rv4 = 32 + 22

= 13 =3.61

Adding and subtracting vectors

ru =2i + 9 jrv=5i + 7.2 j

ru+

rv=7i +16.2 j

ru−

rv=−3i +1.8 j

Add and subtract

i

j rv1 =2i

rv3 =−i −2 j

rv2 =i −1.5 j

rv4 =−3i + 2 j

Find:rv1 +

rv2 ,

rv1 +

rv2 ,

rv4 −

rv3

rv1 =2i

rv3 =−i −2 j

rv4 =−3i + 2 j

Find rv1 +

rv2 ,

rv1 +

rv2 ,

rv4 −

rv3

Solution

rv2 =i −1.5 j

Very important!!!rv1 +

rv2 ≠

rv1 +

rv2

rv4 −

rv3 ≠

rv4 −

rv3

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rv 1 +

r v 2 = 3ˆ i −1.5 ˆ j

r v 1 +

r v 2 = (3)2 + (−1.5)2 = 32 +1.52 = 3.35

r v 4 −

r v 3 = (−3ˆ i + 2 ˆ j ) − (−ˆ i − 2 ˆ j )

= −3ˆ i + 2 ˆ j + ˆ i + 2 ˆ j = −2ˆ i + 4 ˆ j

r v 4 −

r v 3 = (−2)2 + (4)2 = 22 + 42 = 4.47

Notations

The following are all common notations of a vector:rv,v,v

The following are common notations of the magnitude

(i.e. the length of the arrow):

v,rv

Vector Components

5

4

-3 i

j

vv =4i −3 j

Terminology

vv =4i −3 j

The x-component of rv is 4, not 4i .

The y-component of rv is -3, not -3 j.

Usually written as:

vx =4,vy =−3

Decomposing a vector

rv

Given rv =7, θ =40o,

what are the x,y components of rv?

vx

vy

It means finding vx ,vy !

Hint: Once you know one side of a right-angle triangle and one other angle, you can find all the lengths using cos, sin or tan.

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θ

A quick reminder

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θ

€

h cosθ€

h sinθ

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h

Trigonometry

cosθ =bh

sinθ =ah

tanθ =ab

Solution

rv =7

Given rv =7, θ =40o,

what are the x,y components of rv?

vx =rv cosθ

=7cos40o =5.36

vy =rv sinθ

=7sin40o =4.50

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θ

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⇒ r

v = 5.36ˆ i + 4.5 ˆ j

Check

On the other hand, given rv =5.36i + 4.50 j,

you can deduce rv and θ.

rv

vx

vy

rv = 5.362 + 4.52 =7 (as expected)

tanθ =oppositeadjacent

=vy

vx

=4.55.36

=0.8396

⇒ θ =tan−1(0.8396) =40o (as expected)

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θ

Angles of a vector

x

Find the angles the four vectors make with the positive x-axis.

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θ1 = 30o

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θ2 = 180o − 30o = 150o

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θ3 = 180o + 30o = 210o or θ3 = −(180o − 30o) = −150o

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θ4 = 360o − 30o = 330o or θ4 = −30o

y

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rv 1

€

rv 4

€

rv 2

€

rv 3

30°

Calculating the angles

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rv = aˆ i + bˆ j

⇒ θ = tan−1(b

a) + φ ,where φ =

0o if a > 0

180o if a < 0

⎧ ⎨ ⎩

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Examples :

r v = 2ˆ i + 3ˆ j ⇒ θ = tan−1(

3

2) + 0o = 56.3o

r v = −2ˆ i + 3ˆ j ⇒ θ = tan−1(

3

−2) +180o = 123.7o

r v = −2ˆ i − 3ˆ j ⇒ θ = tan−1(

−3

−2) +180o = 236.3o

r v = 2ˆ i − 3ˆ j ⇒ θ = tan−1(

−3

2) + 0o = −56.3o(= 360o − 56.3o = 303.7o)

Write down the following three vectors in i j notation. Find the sum of these vectors also.

60o

50o

10o

5 4

4.5

rv1 =5cos60oi + 5sin60o j

=2.5i + 4.3 j

rv2 =4sin50oi −4cos50o j

=3.1i −2.6 j

rv3 =−4.5cos10oi −4.5sin10o j

=−4.4i −0.8 j

rvtotal =1.2i + 0.9 j

(-1) times a vector?

5

4

3 i

j vv =4i + 3 j

−i

−j

What is -vv ?

vv =4i + 3 j ⇒ −vv=−4i −3 jPoints in the OPPOSITE direction!

What is -vv ?

5

4

3 vv =4i + 3 j

5

4

3 −vv = −4i − 3 j

In General

If this is rv

This is -rv

Adding Vectors Diagrammatically

ru

rv

ru +

rv

rv

You are allowed to move an arrow around as long as you do not change its direction and length.

Method for adding vectors:

1. Move the arrows until the tail of one arrow is at the tip of the other arrow.

2. Trace out the resultant arrow.

Addition of vectorsYou are allowed to move an arrow around as long as you do not change its direction.

Adding in a different order

Order does not matter

Subtracting Vectors Diagrammatically

ru

rv

ru

rv

−rv

ru +

rv

ru −

rv

What about ru −

rv?

rv

ru −

rv=

ru+ (−

rv)

If we know (−rv) we know

ru−

rv

Example

Example

What is rA +

rB+

rC?

rA

rC

rB

rD

rA +

rB+

rC

rA +

rB+

rC =−

rD

Adding vectors 1Add the three vectors to find the total displacement.

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rs 1 = 2.6 ˆ j r s 2 = 4ˆ i

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rs 3 = 3.1cos(45o)ˆ i + 3.1sin(45o) ˆ j = 2.19ˆ i + 2.19 ˆ j

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rs total =

r s 1 +

r s 2 +

r s 3 = (4 + 2.19)ˆ i + (2.6 + 2.19) ˆ j = 6.19ˆ i + 4.79 ˆ j

or more precisely :r s total = (6.19ˆ i + 4.79 ˆ j )km

Adding Vectors 2

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Find r A − 2

r B .

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rA = 2.8cos(60o)ˆ i + 2.8sin(60o) ˆ j = 1.40ˆ i + 2.42 ˆ j r B = 1.9cos(60o)ˆ i −1.9sin(60o) ˆ j = 0.95ˆ i −1.65 ˆ j

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2r B = 2(0.95ˆ i −1.65 ˆ j ) = 1.90ˆ i − 3.30 ˆ j

⇒r A − 2

r B = (1.40 −1.90)ˆ i + (2.42 + 3.30) ˆ j

= −0.50ˆ i + 5.72 ˆ j

Electric field is a vector

The direction of the electric field is given below:

Vector Notation of E field

r : Unit vector pointing from the charge to the observer.

r =1

q

r

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rE =

q

4πε 0r2

ˆ r

Charges produce electric field. The closer you are to the charge, the stronger is the electric field. Unit: V/m = N/C

ε0 = 8.85 ×10−12 C 2N −1m−2 : Permittivity

Another Notation

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rE =

q

4πε 0r2

ˆ r =1

4πε 0

q

r2ˆ r =

kq

r2ˆ r

k =1

4πε0

=8.99 ×109 Nm2C−2

Store k in your calculator

Type:“8.99E9” then “STO” then “ALPHA” then “K” then “ENTER”

If q=2C, r=1.3m, to find the E field, type:“K*2/1.32”

Electric Field (Magnitude)The magnitude of the electric field produced by a single point charge q is give by:

rE =

q4πε0r

2

Don’t forget the absolute value!Magnitude is always positive.

Warnings

Do not confuse the displacement vector r with rE.

r (blue arrow) points from the charge to the observer.rE (red arrow) is ALWAYS drawn with its tail at the obeserver,

and can point either away from OR toward the charge.

+Observerr

rE

− r rE

Direction of E (one charge)

If q > 0, rE is in the SAME direction as r.

If q< 0, rE is in the OPPOSITE direction as r.

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rE =

q

4πε 0r2

ˆ r

What is r?

Example

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q1 = −1nC

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Observer at P.r r P1 is the displacement vector from q1 to P.

ˆ r P1 is the corresponding unit vector.

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rr P1 = 1ˆ i − 0.5 ˆ j r r P1 = 12 + 0.52 = 1.12

ˆ r P1 =1ˆ i − 0.5 ˆ j

1.12= 0.89ˆ i − 0.45 ˆ j

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ˆ r P1 =r r P1r r P1

Example (Continued)

The electric field vector is given by the red arrow.

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q1 = −1nC

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rE 1 =

q1

4πε 0rP12

ˆ r P1

=−10−9

4πε 0(1.12)2 (0.89ˆ i − 0.45 ˆ j )

= (−6.38ˆ i + 3.22 ˆ j )V /m

The strategy in finding the electric field vector

1. Draw an arrow from the charge to the observer

2. Write down the vector rr and its magnitude

rr

3. Calculate r =rrrr

4. Calculate rE =

q4πε0r

2 r

5. If there are more than one charge, repeat for each one

6. rEtotal =

rE1 +

rE2 +L +

rEN

Find the unit vectors

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A

€

B

€

C

€

D

€

E€

q1

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Find :r r A1r r A1

ˆ r A1

then do the same

for the other points.

Warnings

Do not confuse the displacement vector rr with

rE.

rr (blue arrow) points from the charge to the observer, and is

used to calculate r.rE (red arrow) is ALWAYS drawn with its tail at the obeserver.

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q1 = −1nC

Example - Two Charges

See supplementary notes

-1nC

+1nC

More about r:Sometimes it is easier to find r using angles, for example:

θ θθ

r =cosθ i + sinθ j

r =−cosθ i + sinθ j

r =−sinθ i + cosθ j

Example

4 cm 3 cm

q1 q2P

q1 =1×10−8C, q2 =−2×10−8C

Find the E field at point P

Solution

4 cm 3 cm

q1 q2P

q1 =1×10−8C, q2 =−2×10−8C

rE1 =

q1

4πε0r12 r1 =

q1

4πε0r12 i

=(1×10−8 )

4π(8.85 ×10−12 )(0.04)2i =(56.2i )kV / m

rE1

rE2

rE2 =

q2

4πε0r22 r2 =

q2

4πε0r22 (−i )

=(−2 ×10−8 )

4π(8.85 ×10−12 )(0.03)2(−i ) =(199.8i )kV / m

r1 r2

r1 =i , r2 =−i

⇒rEtotal =

rE1 +

rE2 = (56.2i +199.8i )kV / m = (256.0i )kV / m

Example

x 7-x

q1 q2P

q1 =1×10−8C, q2 =2×10−8C

Find the point P such that E = 0.

7 cm

Example

x 7-x

q1 q2P

q1 =1×10−8C, q2 =2×10−8C

q1

4πε0x2 =

q2

4πε0 (7 −x)2⇒

q1

x2 =q2

(7 −x)2

⇒(7 −x)2

x2 =q2

q1

⇒7−x

x=±

q2

q1

=± 2

⇒ 7−x=± 2x

⇒ x=7

1± 2=−16.9cm (rejected) or 2.9cm

The difference between field vectors and field

linesField vectors Field lines

Properties of field lines

• Field lines never cross each other• Field lines never terminate in vacuum• Field lines originate from positive charges and terminate at negative charge

• Field lines may go off to infinity• The tangent of a field line gives the direction of the E field at that particular point

Dipole

Field vectors Field lines

Similar to this

You connects the field vectors to find the field lines.

Electric Field and Electric Force

Electric field can be used to calculate the electric force:

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rF = q

r E

F and E are parallel when q is positive.F and E are opposite when q is negative.

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Example :

A charge q = −2C in an electric field r E = (2ˆ i − 3ˆ j )N /C will feel a force :

r F = (−2C)(2ˆ i − 3ˆ j )N /C = (−4ˆ i + 6 ˆ j )N

Two point chargesq1 q2

rE21E12

The E field (magnitude) at

point 2 due to charge 1:

E21 =q1

4πε0r2

The force (magnitude) on

charge 2 due to charge 1:

F21 = q2E21 =q1q2

4πε0r2

The E field (magnitude) at

point 1 due to charge 2:

E12 =q2

4πε0r2

The force (magnitude) on

charge 1 due to charge 2:

F12 = q1E12 =q1q2

4πε0r2

Coulomb’s Law

F =q1q2

4πε0r2

The mutual force due to two point charges has magnitude:

Another Notation

F =q1q2

4πε0r2

⇔ F = kq1q2

r2

k =1

4πε0

=8.99 ×109 Nm2C−2

Finding the Electric Force

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q1

€

q2

€

q3

There are two (equivalent) methods of finding the force on a charge (say, q1).

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Method 1 (using E field - recommended) :

Find the electric field at point 1 due to the other two chargesr E 1 =

r E 12 +

r E 13 (Field at point 1 due to q2 and q3)

⇒r F 1 = q1

r E 1

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Method 2 (using Coulomb's Law) :

Find the force vectors on charge 1 due to the other two charges using Coulomb's Lawr F 1 =

r F 12 +

r F 13 (Forces at point 1 due to q2 and q3)

Note that you must first write the forces as vectors, cannot add the magnitude insteadr F 1 ≠

r F 12 +

r F 13

Finding the force on a charge

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At point P :r E = (−9.60ˆ i − 3.16 ˆ j )N /C

P

-1nC

+1nC

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If a charge q3 = −2nC is placed at point P,

the force on it will be given by :r F 3 = q3

r E = (−2nC)(−9.60ˆ i − 3.16 ˆ j )N /C

= (−19.20ˆ i − 6.32 ˆ j ) ×10−9 N