1
PROBLEM 1
a. 3rd Newton’s law states that the force from one on another and vice versa should have the same magnitude and
opposite direction.
b. Let their charges be q1 and q2 , so q1 + q2 = Q = 90.0 µC. So, q2 = Q − q1 and the force between them will be
F =
1 q1 (Q−q1 )
,
4π0
r2
(1)
where r = 1.16 m and F = 12.0 N .
Solving this equation w.r.t. q1 yields
q1 =
Q±
√
Q2 −4F (4π0 )r 2
,
2
(2)
these two solutions correspond to the fact that√
first and second charges may be flipped √
without changing the magnitude of the force. So we may choose q1 =
Q+
Q2 −4F (4π0 )r 2
2
= 60.2 µC and q2 =
Q−
Q2 −4F (4π0 )r 2
2
= 29.8 µC.
c. 3rd Newton law still holds.
d. If the force is attractive, the charges need to be opposite signs. Equation (1) will change as now q1 q2 < 0, and
F → Fneg = −12.0 N . Solution (2) for q1 does not depend on the sign of F , so now
q1 =
Q±
√
Q2 −4Fneg (4π0 )r 2
2
=
Q±
√
Q2 +4F (4π0 )r 2
.
2
(3)
√
Note that these two solutions for q1 have opposite signs and correspond to q1 and q2 : q1 =
√
Q− Q2 +4F (4π0 )r 2
106.8 µC and q2 =
= −16.8 µC.
2
Q+
Q2 +4F (4π0 )r 2
2
=
PROBLEM 2
a. Forces from charges in the corners are same in magnitude. Forces from positive charges +q are directed 45◦ from
the vertical axis down to the left negative charge and to the right negative charge. Forces due to the negative
charges are directed in the same way. So, total horizontal component of the force acting on +Q is zero, which
is also understandable from symmetry arguments: we can flip the picture around vertical axis containing the
charge +Q.
√
qQ
1
√
The vertical component is Fy = −4 sin 45◦ 4π
= − π20 qQ
d2 .
0 (d/ 2)2
√
~ = F~ /Q, so Ex = 0, Ey = − 2 q2 .
b. Electric field E
π0 d
c. The electric field at the center does not change, so if Q → −Q, then F → −F : Fy =
√
2 qQ
π0 d2 ,
Fx = 0.
d. Remains the same.
PROBLEM 3
Dipole moment p~ in this problem corresponds to the initial orientation of the dipole.
a. ∆U = Uf inal − Uinitial = −pE cos 90◦ − (−pE cos 180◦ ) = 0 + pE(−1) = −pE.
b. The change in the potential energy is negative, so the work done by the electric field is positive, Wf ield = −∆U .
Observer counteracts this work because total work is zero, as the change in kinetic energy is zero: ∆Ek =
Wtotal = Wf ield + Wobserver = 0, so her work is negative: Wobserver = −Wf ield = ∆U = −pE.
2
PROBLEM 4
~ = −|e|E,
~ because electron is negatively charged particle. So the force is
a. The force on the electron is F~ = eE
directed against the field and as long as the velocity is parallel to the field, against its velocity. So, acceleration
is directed opposite to velocity, which means that the electron is decelerating.
b. Velocity and acceleration of the electron are along one line, so electron can’t leave this line. It’s linear motion
with constant acceleration. Electron will continue its motion forward until it stops and starts moving backwards
indefinitely.
2
c. vf2 inal = vinitital
+ 2a∆x, where vf inal = 0, vinitial = v, a =
∆x =
v2
2|a|
=
mv 2
2|e|E
F
m
=
−|e|E
m ,
so
= 0.189 m = 18.9 cm.
d. When electron returns to its starting point, its velocity is −vinitial due to total energy conservation. ∆t =
2vm
−8
= 27.5 ns.
|e|E = 2.75 × 10
(4)
∆v
a
=
PROBLEM 5
The charge is placed symmetrically at the center of the cube, so the flux through each of the cube’s faces is equal.
To see this, note, that we can rotate the cube around the center so that any face coincides with any other face after
rotation. This proves that the fluxes through each of the faces are equal, so it should be equals to Φ6 , as there are
total 6 faces. By the way, Φ = − q0 , so the flux through one of the faces is − 6q0 .