Slide 1

Report
Electric Field
Questions from
2004 and 2005
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• The diagram shows two particles at a distance d
apart. One particle has charge +Q and the other -2Q.
• The two particles exert an electrostatic force of
attraction, F, on each other. Each particle is then
given an additional charge +Q and their separation is
increased to a distance of 2d.
• What is the force that now acts between the two
particles?
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• Distance doubles so force drops to a quarter
of that before
• Product of charges before was
+Q x -2Q = -2Q2
• Product of new charges is
2Q x -Q = -2Q2
• So force doesn’t change in sign or size due to
charge change
• Overall change produces an attractive force
of F/4
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The electrical field strength, E, and the electrical potential, V, at
the surface of a sphere of radius r carrying a charge Q are given by
the same equations as that for a point charge on your data sheet.
• A school van de Graaff generator has a
dome of radius 100 mm. Charge begins
to leak into the air from the dome when
the electric field strength at its
surface is approximately 3 × 106 Vm-1.
• What, approximately, is the maximum
potential to which the dome can be
raised without leakage?
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V = Er
100mm = 0.1m
So V = 3 × 106 Vm-1 x 0.1m
= 3 × 105 V
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• At a distance R from a fixed charge,
the electric field strength is E and the
electric potential is V.
• What is the electric field strength and
electric potential at a distance 2R from
the charge?
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• Field strength is an inverse square
relationship therefore doubling the
distance will reduce E to E/4
• Potential is an inverse relationship –
doubling the distance will halve the
potential V to V/2
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• Two charges, P and Q, are 100 mm
apart. X is a point on the line between P
and Q.
• If the potential at X is 0V, what is the
distance from P to X?
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At X the positive electric potential from the +4mC
charge is equal to the negative potential from the 6mC charge
– then when you add them they will cancel out!
Let r be the distance PX
V = constant x Q/r
At X: 0 = constant (4/r – 6/(100-r)
4/r = 6/(100-r)
400 - 4r = 6r
10r = 400
r = 40mm
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• Two isolated point charges are
separated by 0.04 m and attract each
other with a force of 20 μN. If the
distance between them is increased by
0.04 m, what is the new force of
attraction?
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• Distance is doubled therefore force
drops to a quarter of what it was
before (inverse square law)
20/4 = 5 μN
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• The diagram shows a
uniform electric field of
strength 10Vm-1
• A charge of 4 μC is
moved from P to Q and
then from Q to R. If
the distance PQ is 2m
and QR is 3m, what is
the change in potential
energy of the charge
when it is moved from P
to R?
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• Moving from P to Q no work is done
(perp to line of action of the force!) so
we ignore that!
• Moving from Q to R work is done:
DV = 10 V/m x 3 m = 30 V
Work done = QDV = 30 V x 4 μC
= 120 μJ
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