Electrostatics & Electric Fields
K Warne
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Coulomb’s Law:Coulomb’s Law:
F =
kQ1Q
2
r2
F
Q2
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
k = coulomb’s constant
= 9 x 109 N. m2.C-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
Increasing the
charge on any
one of the
spheres will
increase the
force by a
proportional
amount.
Increasing the
charge on any
one of the
spheres will
increase the
force by a
proportional
amount.
F =
kQ1Q
2
r2
F
Q2
Q1
1.
F2=
k2Q1Q
2
r2
2F
Q2
2Q12.
F2=2F
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7.1 ELECTRIC 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
direction of the ______________ to that line.
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7.2 ELECTRIC FIELD PATTERNS:
- +
++
VERY SMALL POINT CHARGES NEAR ONE ANOTHER
- +
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+ + + + + + + + + + +
- - - - - - - - - - -
Between oppositely charged plates
Field is _______________ the oppositely charged plates.
(Force experienced by a charge placed anywhere between the plates is
___________________)
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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 ____________)
+ F
E
Eg: What force would be experienced by an electron in an electric field of 1 x 10-6 NC-1?
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q F
E
Q E =
Fq
but F =
kQ1Q
2
r2
so E =q
kQq
r2
so E = kQ
r2
r
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Electric field due to multiple charges
Calculate the electric field strength at X.
X
B
-2C
A
+2C
1cm2cm
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) SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY
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Work done W = F x d
but F = QE (Def of E)
W = QEd
If the charge is pushed to the left the work done on
the charge is: W = QEd
If the charge is now released, it moves spontaneously to the right,because the field does work on the charge:
W = QEd
Kinetic energy gained = work done
1/2
mv2 = QEd => v = √2QEd/m
Consider applying a force F needed to move a charge from A to B. The charge moves a distance d. The size of
the charge is Q.
7.4 WORK DONE IN A UNIFORM ELECTRIC FIELD
+ d -
+ -
+ -
+ B A -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
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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.
E = 1.5x106 N.C-1 d = 2 nm = 2 x 10-9 m m = 1.7 x 10-27 kg
Q = 1.6 x 10-19 C v = ?
Formula?
v = √2QEd/m
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• A positive charge moves 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 Ek IN ELECTRIC FIELDS
+ -
+
-
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Energy in a radial field
Work must be done on Q2 to move it across 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.
+Q
2
Q1
•r
Energy = Work done
= F x d
= kQ1
Q2
x r
r2
U = kQ1
Q2
r
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.
POTENTIAL DIFFERENCE (V)
+
B A
Potential difference =
Unit: ………………………………….
The potential difference between two points in an electric field is the
………………………………………………………. in moving the charge from the one point to the other.
Define the electric potential at a point as the …………………………… per ………………, i.e. the potential energy
…………………………. would have if it were placed …………………...
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Electric Potential
+Q
2
Q1
•r
= kQ1
Q2
r
Q2
= kQ1
r
Charge at that point
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ -
ELECTRIC FIELD BETWEEN 2 PARALLEL PLATES
If plates are s apart and p.d across plates is V, then
V = W = QEd = EdQ QE = V
d whereE = electric field strength in Vm-1
V = potential difference in V andd = 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]
V
d
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Milikan’s Experiment
• These droplets are charged as they are forced out of the nozzle.
• Milikan used a microscope to observe the oil droplets between the plates.
• As the plate voltage increases some of the drops fall more and more slowly until the drops stop moving.
• At this point the electric force is equal to the weight of the oil droplet.
V = Ed = (F/q).d .: F = q.V/d Felec
= Fg
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.
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Hi -
This is a SAMPLE presentation only.
My FULL presentations, which contain loads more slides and other resources, are freely available on my resource sharing website:
www.sciencecafe.org.za
(paste into your browser if link above does not work)Have a look and enjoy!
Keith Warne