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# G11 Electrostatics & Electric Fields

## by Science Café, Science Teacher at Bishops Diocesan College on May 12, 2010

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A set of slides created to teach G11 Electrostatics & Electric Fields to learners at Bishops Diocesan College in Cape Town.

A set of slides created to teach G11 Electrostatics & Electric Fields to learners at Bishops Diocesan College in Cape Town.

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## G11 Electrostatics & Electric FieldsPresentation Transcript

• G11 Electrostatics
K Warne
• k…1…..2
r…
F =
Coulomb’s Law:
The electrostatic force between two point charges is ...........................to the product of the charges and ................ proportional to the ........................of the distance between them.
Q1
Q2
F
k = .................. constant
= 9 x 109 .... . ....2. ....-2
Calculate the force between an electron and a proton if the distance between them is 1nm.
• kQ1Q2
r2
F =
Coulomb’s Law:
The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Q1
Q2
F
k = coulomb’s constant
= 9 x 109 N.C2.m-2
Calculate the force between an electron and a proton if the distance between them is 1nm. (e- = -1.6x10-19 )
F = kQ1Q2/r2 = (9x109)(-1.6x10-19 )(1.6x10-19)/(1x10-9)2 = -2.304x10-10N
• kQ1Q2
r2
F =
Force and charge
Increasing the charge on any one of the spheres will ………………… the force by a ………………… amount.
Q2
Q1
1.
F
…Q1
2.
Q2
….F
F2=
F2=……
• kQ1Q2
r2
F =
Force and charge
Increasing the charge on any one of the spheres will increase the force by a proportional amount.
Q2
Q1
1.
F
2Q1
k2Q1Q2
r2
2.
Q2
2F
F2=
F2=2F
• Force and charge
Increasing the ……………… between the spheres will
………………… the force by the
…………………… of the proportional amount .
Q1
kQ1Q2
r2
1.
Q2
F=
F
Q1
Q2
2.
1/4F
F2 =
…..
F2=………
• kQ1Q2
(2r)2
F2 =
Force and charge
Increasing the distance between the spheres will decrease the force by the SQUARE of the proportional amount .
Q1
kQ1Q2
r2
1.
Q2
F=
F
Q2
Q1
2.
1/4F
2r
F2=1/4F
• 7.1ELECTRIC FIELD
A region ____________ in which a charge will experience a “_________” or electrostatic
_______________________.
+
-
+
ELECTRIC FIELD LINE:
A line drawn in such a way that at at any point on the line, a small ___________ charge will experience a ___________ in the directionof the ______________ to that line.
+
-
• 7.1ELECTRIC FIELD
Repelled by positive plate
- attracted by negative plate.
A region in space in which a charge will experience a “coulomb” or electrostatic force of attraction or repulsion.
+
-
+
ELECTRIC FIELD LINE:
A line drawn in such a way that at any point on the line, a small positive charge will experience a force in the direction of the tangent to that line.
This would occur at any point between the plates
+
-
and so we can represent the field by a series of lines.
• +
-
PROPERTIES OF ELECTRIC FIELD LINES
Stronger the field - _____________field lines
Equally spaced parallel lines - ________ field.
(i.e. same force experienced at all points)
Field lines meet an object at _______________
The field is continuous and _____________
Field lines never ______ or ___________ because they represent _____________ forces.
• +
-
PROPERTIES OF ELECTRIC FIELD LINES
Stronger the field - closer field lines
Equally spaced parallel lines - uniform field
(i.e. same force experienced at all points)
Field lines meet an object at 90
The field is continuous and is 3 dimensional.
Field lines never cross or intersect because they represent resultant forces.
• 7.2 ELECTRIC FIELD PATTERNS:
VERY SMALL POINT CHARGES NEAR ONE ANOTHER
+
-
+
+
-
+

• ELECTRIC FIELD PATTERNS:
VERY SMALL POINT CHARGES NEAR ONE ANOTHER
• +
-
+
+
-
+
7.2 ELECTRIC FIELD PATTERNS:
VERY SMALL POINT CHARGES NEAR ONE ANOTHER
• Between oppositely charged plates
+ + + + + + + + + + +
- - - - - - - - - - -
Field is _______________ the oppositely charged plates.
(Force experienced by a charge placed anywhere between the plates is ___________________)
• Between oppositely charged plates
+ + + + + + + + + + +
- - - - - - - - - - -
Field is uniform between the oppositely charged plates.
(Force experienced by a charge placed anywhere between the plates is identical)
• ELECTRIC FIELD STRENGTH (E)
For an electric field E = /
where E = ______________
strength in ______
F = _______ in___
Q = _________ in ___
NB. Electric field strength is a ___________quantity (direction: _____________ to ____________)
E
+
F
Eg: What force would be experienced by an electron in an electric field of 1 x 10-6 NC-1?
• ELECTRIC FIELD STRENGTH (E)
For an electric field E = F/Q
where E = electric field
strength in NC-1
F = force in N
Q = charge in C
NB. Electric field strength is a vector quantity (direction positive to negative)
E
F
+
Eg: The force experienced by an electron in an electric field of 1 x 10-6 NC-1?
F = EQ = (1 x 10-6)(-1.6 x 10-19 ) = -1.6x10-25 N .: F = 1.6x10-25 N opposite direction to the field.
• FIELD AROUND A SMALL CHARGED OBJECT
Strength of field at A is:
E =
where:
E = …………………………………
k = …………... constant = …………………
Q = ……. charge [C]
r = ……………………..
Example: Find the electric field strength 2 cm from a proton.
Q

A
+
r
……………………………………………………………………
……………………………………………………………………
• Q

A
+
r
FIELD AROUND A SMALL CHARGED OBJECT
Strength of field at A is:
E = kQ
r2
where:
E = electric field strength at A
k = coulomb’s constant = 9 x 109 N.m2.C-2
Q = + charge [C]
r = distance (m) to A
Example: Find the electric field strength 2 cm from a proton.
E = kQ = 9x109x 1,6 x 10-19 = 3,6 x 10-6N.C-1 away
r2(2 x 10-2)2
• Electric field due to multiple charges
A
+2C
B
-2C
2cm
1cm
X
Calculate the electric field strength at X.
Field due to A:
E =
Field due to B:
E =
• Electric field due to multiple charges
A
+2C
B
-2C
2cm
1cm
X
Calculate the electric field strength at X.
Field due to A:
E = kQ1 /r2
=(9x109)(2x10-6)/(2x10-2)2
= 4.5 x 107 N.C-1 away from A
Field due to B:
E = kQ1 /r2
=(9x109)(-2x10-6)/(1x10-2)2
= -1.8x108 N.C-1
= 1.8x108 N.C-1 towards B
E = E1 + E 2 = 4.5 x 107 + ( 1.8x108 ) = 2.25x108 N.C-1
E = 2.25x108 N.C-1 TOWARDS B (AWAY FROM A)
• + s -
+ -
+ -
+ BA -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD
Consider applying a force F needed to …………………………………. The charge moves a ………………………… The size of the charge is ……..
Work done W = ………..
but F = ……….
W = ……….
If the charge is pushed to the left the …………………. on
the charge is: W = …………
If the charge is now released, it moves spontaneously to the …………, because the …………… does …………………. on the charge:
W = QEs
Kinetic energy gained = ……………….
…………………………………….
• + s -
+ -
+ -
+ BA -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD
Consider applying a force F needed to move a chargefrom A to B. The charge moves a distance s. The size of the charge is Q.
Work done W = F x s
but F = QE (Def of E)
W = QEs
If the charge is pushed to the left the work done on
the charge is: W = QEs
If the charge is now released, it moves spontaneously to the right,because the field does work on the charge:
W = QEs
Kinetic energy gained = work done
1/2mv2 = QEs => v = √2QEs/m
• + A -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ B -
MOVING at 900 to A UNIFORM ELECTRIC FIELD
If a charge is moved at …….. to the field from A to B,
……………… is done on the charge!
+
The force needed……………………………… is ………. N.
Therefore …………………….. !
• + A -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ B -
MOVING at 900 to A UNIFORM ELECTRIC FIELD
If a charge is moved at 90 to the field from A to B,
no work is done on the charge!
+
+
The force neededin the direction of motion is 0 N.
Therefore no work done!
• Question
A proton is accelerated by an electric field of 1.5x106 N.C-1 over a distance of 2 nm. The mass of a proton is 1.7 x 10-27 kg
Calculate the final velocity attained by the proton if it started from rest.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
• Question
A proton is accelerated by an electric field of 1.5x106 N.C-1 over a distance of 2 nm. The mass of a proton is 1.7 x 10-27 kg
Calculate the final velocity attained by the proton if it started from rest.
Kinetic energy gained = work done
1/2mv2 = QEs
v = √2QEs/m
= √ 2(1.6x10-19) (1.5x106 )(2x10-9)/1.7x10-27
= 751.5 m.s-1
• -
+
7.5 WORK, Ep AND EkIN ELECTRIC FIELDS
• A positive chargemoves spontaneously … ...............................................of the field.
• A ……………………………..moves spontaneously in a direction……………to that of the electric field.
• Thus, at any point in an electric field an electric charge possesses ……………………(………..)
• Where free to move, it will ………………..
• It will therefore gain ………. as it loses …….
ie. …………… = ………………. (ignoring air friction).
……………………… = ……………………
+
-
• -
+
7.5 WORK, Ep AND EkIN ELECTRIC FIELDS
• A positive chargemoves spontaneously in the directionof the field.
• A negative chargemoves spontaneously in a directionoppositeto that of the electric field.
• Thus, at any point in an electric field an electric charge possesses potential energy(Ep)
• This Electrical Potential energy is therefore due to the position of the charge relative to other charges.
• Where free to move, it will accelerate.
• It will therefore gain Ek as it loses Ep
ie. Ek gained = Ep lost (ignoring air friction).
+
-
• Energy in a radial field
Q1
Work must be done on Q2 to move it to a distance r from Q1.
If released Q2 must therefore have the potential to move away again with that same energy.
Q2 therefore has Potential energy U at a distance r from Q1.

Q2
+
r
Energy = Work done
= F x s
= kQ1Q2 x r
r2
U= kQ1Q2
r
• +
POTENTIAL DIFFERENCE (V)
Define the electric potential at a point as the ……………………………per ………………, i.e. the potential energy …………………………. would have if it were placed …………………...
A positive test charge is at a ……………….. …………………. Ep at B and at a ………… at A.
There is thus a potential difference (V) between B and A.
 
B A
The potential difference between two points in an electric field is the ………………………………………………………. in moving the charge from the one point to the other.
Potential difference =
Unit: ………………………………….
• +
POTENTIAL DIFFERENCE (V)
Define the electric potential at a point as the electrical potential energyper unit charge, i.e. the potential energy a positive test charge would have if it were placed at that point.
 
A positive test charge is at a higher potential energy Ep at B and at a lower Ep at A.
There is thus a potential difference (V) between B and A.
B A
The potential difference between two points in an electric field is the work done per coulomb of charge in moving the charge from the one point to the other.
Potential difference = Potential lost/work done V = W
charge moved Q
Unit: volt 1 V =1 J.C-1
• THE VOLT
Definition: The volt: the potential difference between any two points in an electric field is one volt if one joule of work is done in moving one coulomb of charge from one point to the other.
NOTE:Zero Ep exists at an infinite distance away from the charge which is producing the electric field. Absolute Ep values are thus all negative.
• + -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
ELECTRIC FIELD BETWEEN 2 PARALLEL PLATES
If plates are s apart and p.d across plates is V, then
V = W = QEs = Es
Q Q
E = V
s where
E = electric field strength in Vm-1
V = potential difference in V and
s = distance apart in m
NOTE: V.m-1 = N.C-1
[V.m-1 = J.C-1.m-1 = N.m.C-1m-1 = N.C-1]
s
V
• Milikan’s Experiment
• Milikan used a microscope to observe the oil droplets between the plates.
• These droplets are charged as they are forced out of the nozzle.
• As the plate voltage increases some of the drops fall more and more slowly until the drops stop moving.
• At this point the electric fore is equal to the weight of the oil droplet.
• The electric force on the object is equal to the weight.
qV = mg
d
q=Droplet Charge V= Holding Voltage  d= Distance between plates  m=droplet mass g= Acceleration due to gravity.
• From his experiments Milikan determined that the charge on an electron was 1.6×10-19 C.
• By timing how long it takes for a droplet to fall with the plates switched off he could calculate the mass of the droplet.