Transcript Force between charges

Forces between charges
A2 Level Notes
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• This formula is on the data sheet
• It applies to point charges a distance ‘r’ apart but can
be used for any charged objects as long as they are
small in diameter compared to ‘r’ – ‘r’ is then the
distance between their centres.
• The fact that F is proportional to 1/r2 is called the
inverse square law of electrostatics.
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Q1
Calculate the force between two small conducting spheres
with charges +1.0 nC and +9.0 nC whose centres are 30 cm
apart in air.
Is the force attractive or repulsive?
Permittivity of air eAIR = 8.9 pN m-1
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A1
From the data sheet
F = 1.0 x 10-9 x 9.0 x 10-9
4p x 8.9 x 10-12 x (0.30)2
F = 8.9 x 10-7 N (P)
The force is positive therefore it is a
repulsive force.
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Electric Field
In the vicinity of any
charge Q there is a region
within which other
charges will ‘feel the
force’ and be repelled by
it or attracted to it.
That region is called the
electric field of charge
Q.
It is a radial field
(spreads out from a point).
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Electric Field
• We can describe the
strength of that field
by considering the force
that would be
experienced by a tiny
test charge ‘q’ placed at
any point in that field.
• The nearer the charge
is to the centre of the
field the stronger the
force it will experience.
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Electric Intensity E
If we divide the force experienced by the test charge ‘q’ by ‘q’ we
eliminate needing to specify the size of our charge. We call this
‘E’ the electric intensity or the electric field strength.
On your data sheet
On your data sheet
F=Qq
4per2
E=F
q
E=Q
4per2
The unit for E could be NC-1 (but Vm-1 is preferred)
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Radial field
Field lines have arrows showing the direction
in which a positive point charge would move if
placed at that position.
Field lines
get further
apart as you
leave the
source –
therefore
the field gets
weaker
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Uniform field
Here you can see
that the unit for E
can also be Vm-1
Field lines
are parallel –
therefore
the field is
constant in
intensity.
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Q2
Point charges are located
at points A and B as shown
in the diagram.
Calculate the electric field
strength at point C.
e AIR = 8.9 pF m-1
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A2
• Both charges will be having an effect of the field
strength at point C.
• Both charges will be producing a radial field around
themselves.
• We therefore have to consider the components of
the intensity of the field at C from each of the
charges.
• From the data book we have the equation:
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A2
Let us find the magnitude of the field intensity at point
C from each charge.
lECAl = 3.6 x 10-9/(4p x 8.9 x 10-12 x (0.030)2)
lECAl = 3.58 x 104 Vm-1 (P)
lECBl = 3.2 x 10-9/(4p x 8.9 x 10-12 x (0.040)2)
lECBl = 1.79 x 104 Vm-1 (P)
(We are going to use these later so they need to be to one
more sig fig than is given in the question)
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A2
We can now add the
vectors at point C.
We know the size and
direction they act in
– we can then
construct a diagram
to add them
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A2
By Pythagoras
E2 = ECA2 + ECB2
E2 = (3.58 x 104 )2 + (1.79 x 104 )2
lEl = 4.0 x 104 Vm-1 (P)
(Final answer – therefore to 2sf)
Tan a = ECB/ECA
= 1.79 x 104 / 3.58 x 104 = 0.5
a = 26.6o (P)
E = 40 kVm-1 at 27o to CA (or 63o to CB) (P)
(Final answer needs direction as E is a vector)
Garfield Graphics included with kind permission from PAWS Inc. All Rights Reserved.
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