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Write down an equation for the force between two point charges, Q1 and Q2 ,
separated by a distance r
(1)
A speck of dust has a mass of 1.0 × 10–18 kg and carries a charge equal to that of
one electron. Near to the Earth’s surface it experiences a uniform downward electric
field of strength
100 N C–1 and a uniform gravitational field of strength 9.8 N kg–1.
Draw a free-body force diagram for the speck of dust. Label the forces clearly.
Calculate the magnitude and direction of the resultant force on the speck of dust.
(6)
(Total 7 marks)
A speck of dust has a mass of 1.0 × 10–18 kg and carries a charge equal
but opposite to that of one electron. Near to the Earth’s surface it
experiences a uniform downward electric field of strength
100 N C–1 and a uniform gravitational field of strength 9.8 N kg–1.
Draw a free-body force diagram for the speck of dust. Label the forces
clearly
mass of 1.0 × 10–18 kg
Charge 1.6 x 10-19C
Fe
Fg
Near to the Earth’s surface
it experiences a uniform
downward electric field of strength
Where Fg is the force due to the gravitational field and Fe
is the force due to the electric field
Calculate the magnitude and direction of the resultant force on the speck of dust.
mass of 1.0 × 10–18 kg
Charge 1.6 x 10-19C
Fg
Fe
Resultant force
Fg+ Fe
Calculate the magnitude and direction of the resultant force on the speck of dust.
mass of 1.0 × 10–18 kg
Charge 1.6 x 10-19C
The force due to the field (Fe)
The force due to gravity (Fg)
F=mg
(where g = 9.8Nkg-1)
From
Fg = 1.0 x 10-18 x 9.8
Fg= 9.8 x 10-18 N
Fe
E
Q
Fe  QE
Fe  1.6 10 19 100  160 10 19 N
Resultant force
Fg+ Fe
The resultant force Fg + Fe= (98 +160) x 10-19 = 258 x 10-19N
The diagram shows a positively charged oil drop held at rest
between two parallel conducting plates A and B.
A
Oil drop
2.50 cm
B
The oil drop has a mass 9.79 x 10–15 kg. The potential difference between the plates
is 5000 V and plate B is at a potential of 0 V. Is plate A positive or negative?
………………………………………………………………………………………………
Draw a labelled free-body force diagram which shows the forces acting on the oil
drop.
(You may ignore upthrust).
(3)
Calculate the electric field strength between the plates.
Electric field strength =…………………………………
(2)
Calculate the magnitude of the charge Q on the oil drop.
Charge =……………………………………
How many electrons would have to be removed from a neutral oil drop for it to
acquire this charge?
………………………………………………………………………………………………
(3)
(Total 8 marks)
The diagram shows a positively charged oil drop held at rest
between two parallel conducting plates A and B.
A
Oil drop
2.50 cm
B
Free body diagram
Fe
Here because the body
is at rest the resultant
force must be zero.
F e = Fg
Fg
The diagram shows a positively charged oil drop held at rest
between two parallel conducting plates A and B.
A
Oil drop
2.50 cm
B
Is plate A positive or negative?
The diagram shows a positively charged oil drop held at rest between two parallel
conducting plates A and B.
A
Oil drop
2.50 cm
B
__ __ __
Fe
+
Fg
Plate A has to have a negative charge
as the drop would accelerate down if
the field were reversed
The potential difference between the plates is 5000 V and plate B is at a potential of 0 V.
Calculate the electric field strength between the plates.
A
Oil drop
2.50 cm
B
The equation for the field strength in a UNIFORM FIELD is
E
V
d
The potential difference between the plates V= 5000V. The distance between the
plates is 2.50 x 10-2m
5 103
5
1
E

2

10
Vm
2.5 102
Calculate the magnitude of the charge Q on the oil drop.
Fe
We want to use
F
E
Q
Fg
We now know that the field has this value
E  2 105Vm 1
The force due to the field is equal to the force of gravity on the
oil drop as it is at rest:
F=mg (where m = 9.79 x 10–15 kg and g =9.81Nkg-1)
F= 9.79 x 10–15 x 9.81 = 9.60 x10-14 N
E
F
Q
F 9.60 1014
19
Q 

4
.
8

10
C
5
E
2 10
This value is 3 electrons
worth of charge