Download Physics 133: Tutorial week 2 Electrostatics

Physics 133: Tutorial week 2
Electrostatics
8. Consider a point 2.0 m away from a −3.0 µC point charge. Calculate
(a) the electric field E and
(b) the electric potential V at this point.
(E = 6750 V m−1 radially inwards; V = −13500 volts)
(a) For a point charge, the magnitude of the electric field is given by
9 × 109 × 3 × 10−6
kq
= 6.75 × 103 N C−1 .
E = 2 =
r
2.02
A positive test charge would experience a force towards the −3 µC charge, hence the field
is radially inwards.
(b) The potential at a distance r from a point charge is given by
kq
9 × 109 × (−3 × 10−6 )
V =
=
= − 13.5 × 103 V .
r
2.0
9. An electron in a uniform electric field experiences a force of 8.0 × 10−16 N. What is
the magnitude and direction of E at this point?
(E = 5000 V m−1 in direction opposite to F)
8.0 × 10−16
F
= − 5 × 103 N C−1 .
=
q
−1.6 × 10−19
The field is in the opposite direction to the force experienced by the electron.
E =
10. Calculate the magnitude of the electric field at the centre of a square 20 cm on a
side if one corner is occupied by a 3.0 µC charge and the other three are occupied by
−2.0 µC charges.
(2.25 × 106 V m−1 )
+3µC
–2µC
20 cm
–2µC
–2µC
At the center of the square, the fields due to the two –2 µC charges at the top right and bottom
left corners are equal and opposite so they cancel. The fields due to the +3 µC charge and the
third –2 µC charge at the centre of the square are in the same direction (indicated by the arrow
in the diagram above). The magnitude of the field at the centre due to the +3 µC charge is
E =
9 × 109 × 3 × 10−6
kq
√
= 13.5 × 105 N C−1 .
=
r2
( 12 2 × 0.2)2
Similarly, the magnitude of the field due to the –2 µC charge at the centre of the square is
E = 9.0 × 105 N C−1 . Since the fields are in the same direction, we must add the contributions.
The field at the center is therefore (13.5 + 9.0) × 105 = 22.5 × 105 N C−1 .
11. A small sphere is given a positive charge and is then brought near a large metal plate.
Draw a diagram showing the electric field lines.
12. A force of 0.032 N is required to move a charge of 40 µC in an electric field between
two points 25 cm apart. What potential difference exists between the two points?
(200 V)
E =
V
d
∴
p.d. = Ed =
F
0.032 × 0.25
= 200 V .
×d =
q
40 × 10−6
13. An electron is accelerated in a machine in which it is subjected to a potential difference of 50 × 106 V. What energy has the electron acquired? Express your answer in
eV.
(50 MeV)
The energy acquired by the electron equals the work done on the electron in moving across the
potential difference. Since p.d. = W/q ,
Energy = qV = 1.6 × 10−19 × 50 × 106 J = 50 × 106 eV = 50 MeV .
14. What is the speed of a 350 eV electron? Take me = 9.1 × 10−31 kg.
(1.1 × 107 m s−1 )
The kinetic energy of the electron KE = 21 mv 2 = 350 eV = 1.6 × 10−19 × 350 J . Hence
v =
2 × 1.6 × 10−19 × 350
9.1 × 10−31
21
m s−1 = 1.11 × 107 m s−1
15. What is the acceleration of an electron in a 2200 N C−1 electric field? The mass of
an electron is 9.1 × 10−31 kg.
(3.9 × 1014 m s−2 )
F = ma = Eq
hence
a =
2200 × 1.6 × 10−19
Eq
= 3.87 × 1014 m s−2 .
=
m
9.1 × 10−31
16. What potential difference is needed to give a helium nucleus (q = 3.2 × 10−19 C)
8.0 keV of kinetic energy?
(4000 V)
The energy of the electron equals the work done on the charge, therefore
p.d. =
W
8.0 × 103 × 1.6 × 10−19
= 4000 V .
=
q
2 × 1.6 × 10−19
17. The work done to move a 125 nC charge from point P to point Q is 3.0 × 10−5 J. If
the charge started from rest and had 1.0 × 10−5 J of kinetic energy when it reached
point Q, calculate the potential difference between P and Q.
(160 V)
Some of the work done is converted into the kinetic energy of the charge, the rest is done
against the electric field. We can write
WT OT = ∆KE + Wfield .
The work done against the field is therefore Wfield = 3.0 × 10−5 − 1.0 × 10−5 J = 2.0 × 10−5 J .
Hence
2.0 × 10−5
W
= 160 V .
=
p.d. =
q
125 × 10−9
18. Two parallel plates are 2 cm apart and connected across 120 volts. Find
(a) the electric field E (assumed uniform) in the region between the plates,
(b) the force on an electron between the plates due to the electric field,
(c) the energy gained by an electron (i) in electronvolts (ii) in joules if it travels
freely from one plate to the other,
(d) the ratio of the electric force to the gravitational force on the electron in (c),
(e) the speed of the electron in (c) when it reaches the positive plate.
(Take me = 9.1 × 10−31 kg and e = 1.6 × 10−19 C.)
(6000 V m−1 ; 9.6 × 10−16 N; 120 eV, 1.92 × 10−17 J; 1.1 × 1014 ; 6.5 × 106 m s−1 )
V
120
=
V m−1 = 6000 V m−1 .
d
0.02
F = Eq = 6000 × 1.6 × 10−19 N = 9.6 × 10−16 N .
(i) The potential difference across the plates is 120 V , hence the energy gained by an
electron travelling freely from one plate to the other is 120 eV .
(ii) 120 eV = 120 × 1.6 × 10−19 J = 1.92 × 10−17 J .
The electric force on the electron was calculated in (b). The gravitational force on the
electron F = mg = 9.1 × 10−31 × 9.8 N = 8.92 × 10−30 N . The ratio of the electric force
to the gravitational force is therefore (9.6 × 10−16 ) ÷ (8.92 × 10−30 ) = 1.1 × 1014 .
Assuming the electron started from rest, the energy gained by the electron is its kinetic
energy. Hence from (c) (ii) and using KE = 21 mv 2 we have
(a) E =
(b)
(c)
(d)
(e)
v =
2 × 1.92 × 10−17
9.1 × 10−31
21
m s−1 = 6.5 × 106 m s−1 .
19. A charged particle remains stationary in an upwardly directed field between two horizontal parallel charged plates separated by 2 cm. Calculate the potential difference
V between the plates if the particle has mass 4 × 10−13 kg and charge 2.4 × 10−18 C.
(3.3 × 104 V)
When the particle is stationary, the force due to gravity FG equals the force due to the electric
field FE , where
FG = mg and
V
FE = Eq = q .
d
Equating the above relations and rearranging, we find
mgd
4 × 10−13 × 9.8 × 10−2
= 3.27 × 104 V .
V =
=
q
2.4 × 10−18