Physics 21
Fall, 2011
Solution to HW3
2196
Positive charge
Q
is uniformly distributed around
a semicircle of radius
a
. Find the electric field (magnitude
and direction) at the center of curvature
P
.
x
y
d
θ
dQ =
λ
ds
=
λ
ad
θ
We start with the equation from the equation sheet that
gives the field at the field point
r
in terms of a charge
dQ
at
the charge point
r
′
:
d
E
(
r
) =
1
4
πǫ
0
dQ
(
r

r
′
)

r

r
′

3
The field point
r
(where we want to know
E
) is at the origin,
so
r
= 0, and the charge point (where
dQ
is located) is
r
′
=
a
cos
θ
ˆ
i
+
a
sin
θ
ˆ
j
,
where
θ
is the angle above the
x
axis, so
r

r
′
=

a
cos
θ
ˆ
i

a
sin
θ
ˆ
j
,
and

r

r
′

=
a.
Since we want to add up all the contribution from all
dQ
,
we must relate
dQ
to
dθ
so that we can integrate over the
angle
θ
spanned by the semicircle. The arc length
ds
swept
out by an angle
dθ
is
adθ
, so the charge
dQ
on
ds
is
dQ
=
λds
=
λadθ,
where
λ
is the linear charge density (charge per unit length):
λ
=
Q
πa
.
With all these substitutions, the original equation becomes
d
E
=
1
4
πǫ
0
(
Q
πa
adθ
)(

a
cos
θ
ˆ
i

a
sin
θ
ˆ
j
)
a
3
=

Q
4
π
2
ǫ
0
a
2
(cos
θ
ˆ
i
+ sin
θ
ˆ
j
)
dθ.
We can find the total field at the origin by integrating:
E
=
integraldisplay
d
E
=
integraldisplay
π
0

Q
4
π
2
ǫ
0
a
2
(cos
θ
ˆ
i
+ sin
θ
ˆ
j
)
dθ.
We will look at each component of the integral over
θ
sepa
rately. We see that
integraldisplay
π
0
cos
θdθ
= 0
and
integraldisplay
π
0
sin
θdθ
= 2
,
so we get
E
=

Q
4
π
2
ǫ
0
a
2
(0
ˆ
i
+ 2
ˆ
j
) =

Q
2
π
2
ǫ
0
a
2
ˆ
j
.
From this result we see the
E
has a magnitude of
Q
2
π
2
ǫ
0
a
2
and points in the

y
direction or downward. The
x
compo
nent of the field is zero, as one would expect from symmetry.
21105
Three charges are placed as shown in the figure.
The magnitude of
q
1
is 2.00
µ
C, but its sign and the value of
the charge
q
2
are not known. Charge
q
3
is +4.00
µ
C, and the
net force
F
on
q
3
is entirely in the negative
x
direction. a)
Calculate the magnitude of
q
2
. b) Determine the magnitude
of the net force
F
on
q
3
.
(a) We first determine the sign of the charge
q
1
. We can do
this by thinking about which direction the force will be in
for the different combinations of signs for charges
q
1
and
q
2
.
Since there is no
y
component of the force on
q
3
we know
that
q
1
and
q
2
must have opposite signs. Since the force is
directed in the negative
x
direction we can infer that
q
1
must
be negative and
q
2
must be positive.
To determine
q
2
we calculate the the total force
F
=
F
1on3
+
F
2on3
on
q
3
. Expressions for
F
1on3
and
F
2on3
follow from
the general expression for
F
2on1
on the equation sheet:
F
1on3
=
1
4
πǫ
0
q
3
q
1
(
r
3

r
1
)

r
3

r
1

3
,
F
2on3
1
4
πǫ
0
q
3
q
2
(
r
3

r
2
)

r
3

r
2

3
.
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 Spring '08
 Hickman
 Charge, Work, Electric charge, vy, Hmax

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