Download Physics (2): Problem set 1 solutions

PHYS 104
Physics (2): Problem set 1 solutions
Problem 1: Two identical charges q = +1 nC are located on the x-axis at positions +2 cm and
−2 cm. What is the electric field at the origin (centre between the two point charges)? What is
the electric field at position x = +4 cm? What is the electric force (magnitude and direction) on
a third charge −q placed on the +y-axis a distance 2 cm from the origin?
Solution:
The electric field at the centre between the two identical charges is just zero (by symmetry).
At position x = +4 cm, the electric field is the sum of two components:
E1x = +k
q
,
d21
E2x = +k
q
d22
where d1 = 0.02 m and d2 = 0.06 m are the distances to the point charges, and both fields are in
the same direction (positive x-axis). Thus E = 2.5 × 104 N/C.
The electric force on the negative charge placed on the y-axis a distance 2 cm from the origin
has only a y-component by symmetry (i.e. Fx = 0). The y-component is
Fy = −2 × k
q2
sin 45◦ = −1.59 × 10−5 N
d2
where d is the distance between the negative charge and the other two positive charges. The
negative sign indicates the direction of the force is along the negative y-axis.
y
−
b
F~1
F~2
2 cm
F~
b
45◦
b
+
+
2 cm
x
2 cm
Figure 1: problem 1
Problem 2: A positive charge +q is placed exactly in the middle between two other identical
charges +q. Show that the particle in the middle exhibits a simple harmonic motion if it is displaced from its equilibrium position a distance ǫ ≪ d, where d is the distance between the middle
charge and the other two charges, and find its period. What would happen if the charge was
negatively charged instead?
1
d+ǫ
d−ǫ
net force
b
b
+
+
b
+
ǫ
Figure 2: problem 2
Solution
A positively charged particle in the middle between two other positively charged particles is
displaced a distance ǫ on the positive x-direction, we have the total force on it is:
q2
kq 2
−
k
(d + ǫ)2
(d − ǫ)2
kq 2
kq 2
−
= 2
d (1 + ǫ/d)2 d2 (1 − ǫ/d)2
kq 2 =
(1 + ǫ/d)−2 − (1 − ǫ/d)−2
2
d
kq 2 2
2
(1
−
2ǫ/d)
−
(1
+
2ǫ/d)
+
O(ǫ
/d
)
=
d2
F =
where in the last equality we ignored O(ǫ2 /d2 ) terms as they are much smaller than O(ǫ/d) terms.
We used in this expansion the fact that (1 ± x)n = 1 ± nx for small x. Thus
kq 2 1 − 2ǫ/d − 1 − 2ǫ/d + O(ǫ2 /d2)
2
d
kq 2
=
[−4ǫ/d]
d2
4kq 2
= − 3 ǫ
d
= −Cǫ
F =
where the constant C = 4kq 2 /d3 . This equation looks like Hooks law Fp= −Cx. Hence the
particle will exhibit a simple harmonic motion with angular frequency ω = C/m. The period of
oscillation is T = 2π/ω.
The middle particle is said to be in a stable equilibrium.
Note: If the charge in the middle is negative, then it would be in an stable equilibrium. In
fact it would exponentially drift to the charge to which it is initially closest (like a ball on the top
of a hill).
Problem 3: Find the expression for the electric field a distance r from a long wire carrying a
linear charge distribution +λ.
2
y
~
δE
δEy
b
θ
δEx
√
x2
+
r2
r
δq = λδx
θ
λ
+++++++++++++++++++++++++++++
δx
x
x
Figure 3: problem 3
Solution
We divide the wire into small elements of length dx of charge dq = λdx. By symmetry we can
deduce that the electric field has only a y-component (see figure 3) since the component dEx cancels
against the x-component of the field from the charge at −x. Hence we only need to compute the
y-component, which is given by:
dEy = dE sin θ =
r
dx
kdq
√
= krλ 2
2
+ r x2 + r 2
(x + r 2 )3/2
x2
The total electric field can be found by integrating the above expression over all values of x. If the
wire has length1 2L, we have:
Z L
dx
Ey = krλ
2
2 3/2
−L (x + r )
This integral can be performed by integration. You can check by differentiation that:
Z
dx
x
= √
2
2
3/2
2
(x + r )
r x2 + r 2
Thus we find:
L
x
√
Ey = krλ
r 2 x2 + r 2 −L
2L
1 λ
√
=
4πǫ0 r L2 + r 2
1
1 λ
p
=
2πǫ0 r 1 + r 2 /L2
1
We are measuring the field in a point belonging to the axis of the wire. We shall set L → ∞ at the end of the
calculation for a very long wire.
3
If we are measuring the electric field at a point very close the finite wire (r ≪ L), or equivalently
if the wire is infinitely long, we simplify the result to (r ≪ L):
E=
1 λ
2πǫ0 r
which is exactly the expression we obtained using Gauss’s law.
Problem 4:
A positive charge +q is placed in the cavity within a neutral hallow conducting sphere of inner
radius R1 and outer radius R2 (R1 < R2 ).
1. Draw a diagram representation for the distribution of charges in the sphere.
2. Calculate the magnitude electric field in all regions of the space.
3. What is the total charge on the inner surface of the conductor, and on the outer surface?
4. Draw electric field lines.
5. Calculate the electric potential at every point in space, assuming that the potential reference
is infinity.
The sphere is connected to the ground. Redraw the distribution of charges and electric field
lines. How would the potential differ in this case?
Solution
1. The drawing is shown in figure 4.
2. To compute the electric field we use Gauss’s law. For this we choose a sphere of radius r
which is concentric with the hallow sphere and charge +q. By symmetry the electric field
(if any) is constant over this surface and is perpendicular to it, hence the flux over Gauss’s
surface is just:
Z
Z
Z
~
~
ΦE~ = E.dA = EdA = E dA = EA = E4πr 2
Then by Gauss’s law we have:
ΦE~ =
qenc
ǫ0
Thus:
E4πr 2 =
Therefore:
E=
qenc
ǫ0
qenc 1
4πǫ0 r 2
We distinguish the following cases:
0 < r < R1 : This is the region within the cavity of the conductor. Here the charge enclosed
is just +q, which means:
q 1
E=
4πǫ0 r 2
4
+
+
−
+
−
−
−
+
−
+
+
−
+
−
−
+
+
Figure 4: problem 4
R1 < r < R2 : This is the region within the conductor. Immediately we notice that E = 0
(inside a conductor). This automatically means that qenc = 0, which implies that the
charge that accumulates in the inner surface of the conductor must cancel out the charge
+q. Thus the charge in the inner surface is −q.
r > R2 This is the region outside the sphere. Since the conductor is neutral the charge on
the outer surface must be +q. and thus the total charge enclosed by Gauss’s surface in
this case is just qenc = +q, and we obtain:
E=
q 1
4πǫ0 r 2
The electric field as a function of r is shown in figure 5
3. The total charge on the inner surface of the conductor is just −q, and that on the outer
surface is just +q. Note that even if the shape of the conductor is irregular this property
would still be correct (i.e. charge in the inner surface exactly equals the opposite of the
charge in the outer surface, which equals the charge in the cavity).
4. Field lines are drawn in fig 4.
R
5. The electric potential V is given by V = − Edr. We shall fix the potential to be zero at
infinity. We discuss the following cases:
5
E(r)
R1
r
R2
Figure 5: E as a function of r.
R
r > R2 : We call this region A. Here integrating the expression for E we obtain ( dr/r 2 =
−1/r)
q 1
+ C1
VA =
4πǫ0 r
where the constant of integration is chosen so as to set V = 0 when r → ∞. Clearly
C1 = 0. We shall need the expression for the voltage at the boundary with the outer
surface:
q 1
VA (R2 ) =
4πǫ0 R2
R1 < r < R2 : Here the electric field is just zero, which means the potential is constant (the
constant of integration): VB = C2 . The constant C2 is fixed by the requirement of the
continuity of the electric potential across the boundaries of the regions. Particulary
when r = R2 we must have VA (R2 ) = VB (R2 ), which gives:
VB (r) =
q 1
4πǫ0 R2
which is constant.
0 < r < R1 : We call this region C, and here we again have:
VC =
q 1
+ C3
4πǫ0 r
and once again the constant C3 is fixed by the continuity of the potential across the
boundary at r = R1 . Here: VC (R1 ) = VB (R1 ), which means:
q 1
q 1
+ C3 =
4πǫ0 R1
4πǫ0 R2
Isolating C3 and substituting we find that:
1
1
1
q
+
−
VC (r) =
4πǫ0 r R2 R1
6
V (r)
V0
r
R1
R2
Figure 6: The potential V (r) as a function of r. In the picture V0 =
q
1
4πǫ0 R2
If the sphere is grounded then the positive charges on the outer surface are grounded and we
obtain the distribution and field lines shown in figure 7.
−
−
−
−
−
+
−
−
−
b
Figure 7: The sphere grounded.
The electric field in this case is zero everywhere except for the region 0 < r < R1 , where it has
the same value as before. Hence the potential is constant for r > R1 and is just equal to zero.
However inside the cavity within the conductor it simply becomes:
V =
q 1
+ C4
4πǫ0 r
where the constant C4 is chosen so as the potential at the boundary at r = R1 is a continuous
7
function. Thus we find:
q
V =
4πǫ0
for r < R1 and zero otherwise.
8
1
1
−
r R1