Test 1
1. a) Write the definition of the electric field vector.
b) Which attributes of particles determine their electromagnetic interaction?
c) Describe, in terms of electrostatic force, the electrostatic interaction between
charged particles.
d) Draw the lines of an electric field produced by a dipole.
2. An electron1 enters the region of a uniform electric field produced by two
parallel plates of length L = 10cm as shown in figure 23.25. The strength of the
field is E = 200 N/m and the initial speed of the electron is vi = 3⋅106 m/s.
a) Find the acceleration of the electron while it is in the electric field.
b) Find the time is takes the electron to travel through the field.
c) Determine the deflection angle.
3. Three “point charges” are located at the corners of an equilateral triangle with
0.5 m long side, as shown in Figure P23.7. Calculate2 the net electric force on
the 7 µC “charge”.
4. A nonuniform electric field is given by the expression E(x,y,z) = [ay,bz,cx],
where a,b and c are constants. Determine the electric flux through a rectangular
surface in the xy-plane, extending from x = 0 to x = w and from y = 0 to y = h
5. A thin rod of length L and uniform linear charge density λ lies along the x-axis,
as shown in Figure P23.35. Show that the electric field strength at point P, a
distance y0 from the rod along the perpendicular bisector is given by
E = 2ksinθ0/y0.
1
2
electron: charge -1.6⋅10-19 C; mass 9.11⋅10-31 kg
Coulomb constant: 9⋅109 Nm2 /C2
-1-
r r
r
a)The electric field vector at a certain position r is a vector E(r ) such
that the electric force exerted on "a charge" q (a particle with charge q)
placed at this position would be
r
r
Fel = qE
b) The electric charge and spin determine electromagnetic interactions.
r
c) The electrostatic force F21 , that particle 1 exerts on particle 2
depends on the charges, q1 and q 2 , of the particles, and the relative
position of the particles.
r
kq q
F21 = 12 2 rˆ12
r
d)
+
-
-2-
a) From the definition of the
electric field vector the force
exerted on the electron is
F = −eE = −eEˆj
We can assume that only this
electrostatic force is exerted on
the electron, therefore from
Newton’s
second
law
the
acceleration of the electron is
(
L
-------------------------vi
E
y
+++++++++++++++++++
vf
x
)
F
eE ˆ − 1. 6 ⋅10 −19 C ⋅ 200 N C ˆ
a = = − j=
j = −3. 5 ⋅ 1013 m 2 ˆj
− 31
s
m
m
9.1⋅ 10 kg
(
)
b) Since the acceleration of the electron is in the vertical direction, the
x-component of the position is a linear function of time. The time
required by the electron to pass the electric field is therefore
∆t =
L
L
0.1m
= =
= 3. 33 ⋅10 −8 s
6m
v x v i 3 ⋅ 10 s
c) In a motion with constant acceleration, velocity is a linear function of
time. With the time reference t0 = 0s at the instant when the electron
enters the field, velocity is described by the following funcion
v = v i + at = [v i , 0] + [0, a ]t = [v i , at ]
Using the scalar components of the velocity at the instant when the
electron leaves the field, one can find the deflection angle
3.5 ⋅1013 m s 2 ⋅ 3.33 ⋅ 10− 8 s
at
θ = arctan = arctan
≈ 21°
vi
3 ⋅ 106 m s
θ
-3F31
y
q3 = 7µC
F
F32
c = 0.5m
x
q2 = -4µC
q1 = 2µC
µC particle
by the other two particles. The electric field produced by the 2µC
particle, at the location of the 7µC particle, multiplied by the charge of
the latter gives the first force
kq1q 3
[cos 60°, sin 60°] =
c2
9 ⋅ 109 Nm2 ⋅ 2 ⋅ 10− 6 C ⋅ 7 ⋅ 10− 6 C
F31 = q 3E1 =
=
(
C
)(
)(
(0.5m )
2
)
[cos 60°, sin 60°] = [0.252,0.436]N
Similarly, the second force exerted by the -4µC particle is
kq 2q 3
[− cos 60°, sin 60°] =
2
c
9 ⋅ 109 Nm2 ⋅ − 4 ⋅ 10− 6 C ⋅ 7 ⋅ 10− 6 C
F32 = q 3E 2 =
(
=
C
)(
)(
(0.5m )
2
)
[− cos 60°, sin 60°] = [0.504,−0.872]N
The net electric force on the 7µC is therefore
F = F31 + F32 = [0.752,−0.436]N
-4y
h
E
y
dy
dx
dA
x
x
w
z
The orientation of the considered surface places all the differential
surface elements in the xy-plane (the z-component for all elements has a
zero value). Additionally, all vectors, assigned to these elements, are in
the z-direction
dA = [0,0, dA ] = [0, 0, dxdy ]
The differential flux (through the marked element) is therefore
dΦ = E ⋅ dA = [ay,0, c x] ⋅ [0,0, dxdy ] = c xdxdy
Integration over the surface requires that the limits for x are 0 and w,
while the limits for y are 0 and h.
x2
Φ = ∫ ∫ cxdxdy = c ∫
0 0
0 2
hw
h
w
0
w2 h
cw 2 h
dy = c
∫ dy =
2 0
2
-5-
The electric field of an extended
object requires integration of the
dE y
electric field produced by the
P
individual differential fragments of
θ0 θ
the object. For a linear object the
charge dq of its fragment of length
r
y0
dx depends on the linear charge
density λ.
dq = λdx
x
O
dx
This statement defines the charge
L
density.
A differential fragment can be considered as a particle (“point
charge”). From Coulomb’s law, the electric field at point P, produced
by the (marked) differential piece of the rod is therefore
dq
λdx
λdx
dE = k 2 rˆ = k 2
[
−
sin
θ
,
cos
θ
]
=
k
[− x, y 0 ]
3
2
2 2
r
x + y 20
x +y
(
0
)
Integration of the above expression over the rod yields the electric field
at point P. From the symmetry, one should conclude that only the
y-component of the field is different than zero.
L
L
L
2
2
λydx
x
2kλ sin θ0
2
Ey = ∫ k
=
k
λ
y
=
2
k
λ
y
=
0
0
2
y0
−L
y 20 x 2 + y20 −L
(x 2 + y 20 )3 2
2 L
2
2
y 0   + y 20
 2
Since the other components have zero values, the electric field strength
is
2kλ sin θ 0
E = E 2x + E 2y + E 2z = E y =
y0