Question 14
(a)
State what is meant by electric potential at
a point. [2]
(b)
The centres of two charged metal spheres A and B are separated by a
distance of 44.0 cm, as shown in Fig. 7.1.
Fig. 7.1 (not to scale)
A moveable point P
lies on the line joining the centres of the two spheres. Point P is a distance x
from the centre of sphere A. The variation with distance x
of the electric potential V at point P is shown in
Fig. 7.2.
Fig.
7.2
(i)
Use Fig. 7.2 to state and explain whether the two spheres have
charges of the same, or opposite, sign. [1]
(ii)
A positively-charged particle is at rest on the surface of sphere A.
The particle moves
freely from the surface of sphere A to the surface of sphere B.
1.
Describe qualitatively the variation, if any, with distance x
of the speed of the particle
as it
moves from x
= 12 cm to x = 25 cm
passes through x
= 26 cm
moves from x
= 27 cm to x = 31 cm
reaches x
= 32 cm
[4]
2.
The particle has charge 3.2 × 10-19 C and mass 6.6 × 10-27 kg.
Calculate the maximum
speed of the particle. [2]
[Total: 9]
Reference: Past Exam Paper – March 2018 Paper 42 Q7
Solution:
(a)
The electric potential at a point is the work done per
unit charge in moving a positive charge from infinity to that point.
(b)
(i)
The charges have the same sign as the potential due to
both of them always has the same sign (it is always positive).
(ii)
1.
(from x = 12 cm to x = 25 cm: speed increases) at
decreasing rate
(from x = 27 cm to x = 31 cm: speed decreases) at increasing rate
at x = 26 cm: speed maximum
at 32 cm: speed still decreasing
{The
electric potential energy of the particle is converted to kinetic energy as the
particle moves away from sphere A.
q ΔV = ½ mv2
The kinetic
energy, and thus speed, depends on the change in potential.
From x = 12
cm to x = 25 cm: speed increase at a decreasing rate (as the change in potential
decreases at a decreasing rate)
From x = 27
cm to x = 31 cm: speed decreases at an increasing rate (as the change in potential
increases at an increasing rate)
At x = 26
cm: the speed is maximum (as the change in potential is maximum)
At x = 32
cm: the speed is still decreasing (as the change in potential has decreased)}
2.
{The
electric potential energy is converted into kinetic energy.}
q ΔV = ½ mv2
{Initial potential of particle = 2.14×104 V
The speed is maximum at x = 26 cm. This
corresponds to a potential of 1.43×104 V.}
3.2 × 10-19 × (2.14 – 1.43) × 104 = ½
× 6.6 × 10-27 × v2
v2 = 6.88 × 1011
v = 8.3 × 105 m s-1
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