Problems with Answers for Pre Test
Problem: A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side.
Determine the magnitude and direction of the force on each charge.
We find the magnitudes of the individual forces on the
charge at the upper right corner:
F1 = F2 = kQQ/L2 = kQ2/L2
y
F1
= (9.0109 N · m2/C2)(6.0010–3 C)2/(0.100 m)2
7
L
= 3.2410 N.
+
+
F3 = kQQ/(L√2)2 = kQ2/2L2
Q
Q
= (9.0109 N · m2/C2)(6.0010–3 C)2/2(0.100 m)2
= 1.62107 N.
L
The directions of the forces are determined from the signs
of the charges and are indicated on the diagram. For the
forces on the upper-right charge, we see that the net force
will be along the diagonal. For the net force, we have
Q
Q
F = F1 cos 45° + F2 cos 45° + F3
+
+
= 2(3.24107 N) cos 45° + 1.62107 N
= 6.20107 N along the diagonal, or away from the
center of the square.
From the symmetry, each of the other forces will have the same magnitude and a direction away from
the center: The net force on each charge is 6.20107 N away from the center of the square.
Note that the sum for the three charges is zero.
F3
F2
x
Problem: Given the two charges Q1= Q and Q1=-Q on a distance d=1 m from each other
shown in Figure below, find at what position x is electric field is equal zero? Is the field
zero at any other points, not on the x axes?
Because the charges have opposite signs, the location where the electric
field is zero must be outside the two charges, as shown.
x
P
The fields from the two charges must balance:
–
x
+
2
2
kQ1/(d + x) = kQ2/x ;
Q1
Q2
E2 E1
d
Q/(d + x)2 = Q/2x2, or d + x = ± x√2.
which gives x = d/(√2 – 1), – d/(√2 + 1), or x = d(1 + √2), d(1 – √2),
Because d(1 – √2) is between the charges, the location is
d(1 + √2) from the negative charge, and d(2 + √2) from the positive charge.
Other than at infinity, there is no place, not on the x-axis, where the vectors sum to zero.
Problem: Figure shows a solid metal sphere at the center of a hollow metal sphere. What
is a total charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow
sphere, and (c) the exterior surface of the hollow sphere?
The excess charge on a conductor resides on the outer surface. The field inside, outside, and within the hollow
metal sphere has spherical symmetry.
The figure shows spherical Gaussian surfaces with radii r  a, a  r  b, and r  b. These surfaces match the
symmetry of the charge distribution.
Solve: (a) For r  a, Gauss’s law is
e   E  dA 
Qin
0

Q
0
Notice that the electric field is everywhere perpendicular to the spherical surface. Because of the spherical
symmetry of the charge, the electric field magnitude E is the same at all points on the Gaussian surface. Thus,


e  EAsphere  E 4 r 2 
Q
0
 E
Q
4 0r
2
E
1 Q
rˆ
4 0 r 2
where we made use of the fact that E is directed radially outward. The field depends only on the enclosed charge,
not on the charge on the outer sphere.
For a  r  b, Gauss’s law is
Q
e   E  dA  EAsphere  E 4 r 2  in


0
Here Qin  0 C. It is not Q, because the charge in the cavity polarizes the metal sphere in such a way that E = 0
in the metal. Thus a charge Q moves to the inner surface. Because the hollow sphere has a net charge of 2Q,
the exterior surface now has a charge of 3Q. Thus, the electric field E  0 N/C.
For r  b,
Qin  Qexterior  Qinterior  Qcavity  3Q  (Q)  (Q)  3Q
Gauss’s law applied to the Gaussian surface at r  b yields:


e   E  dA  EAsphere  E 4 r 2 
Qin
0

3Q
0
 E
1  3Q 
1  3Q 
 2  E

 rˆ
4 0  r 
4 0  r 2 
(b) As determined in part (a), the inside surface of the hollow sphere has a charge of Q, and the exterior surface
of the hollow sphere has a charge of 3Q.
Assess: The hollow sphere still has the same charge 2Q as given in the problem, although the sphere is
polarized.
Problem: Four point charges are located at the corners of a square that is 8.0 cm on a side.
The charges, going in rotation around square, are Q, 2Q, -3Q and 2Q, where Q=4.5 µC
(see Figure). What is total electric potential energy stored in the system, relative to U=0
at infinite separation?
For the potential energy of the four charges we have
U = (1/4Å0)(Q1Q2/r12 + Q1Q3/r13 + Q1Q4/r14 + Q2Q3/r23 +
Q2Q4/r24 + Q3Q4/r34)
2
= (Q /4Å0)[(1)(2)/b + (1)(– 3)/b√2 + (1)(2)/b + (2)(–3)/b +
(2)(2)/b√2 + (– 3)(2)/b]
9
2
2
–6
2
= (9.010 N · m /C )(4.810 C) 
[2 – 3/√2 + 2 – 6 + 4/√2 – 6]/(0.080 m)
= – 19 J.
Q +
b
+
2Q
We find the electric potential energy of the system by considering one of the charges to be at the
potential created by the other charge. This will be zero when they are far away. Because the masses
are equal, the speeds will be equal. From energy conservation we have
K + U = 0;
½mv2 + ½mv2 – 0 + Q(0 – V) = 0, or
2(½mv2) = mv2 = Q(kQ/r) = kQ2/r;
(1.010–6 kg)v2 = (9.0109 N · m2/C2)(7.510–6 C)2/(0.055 m),
which gives v = 3.0103 m/s.
Problem: A proton is fired a proton with a speed of 200 000 m/s from the midpoint of the
capacitor toward the positive plate. (a) show that this is insufficient field to reach the
positive plat. (b) What is the proton’s speed as it collides with the negative plate?
Energy is conserved. The proton’s potential energy inside the capacitor can be found from the capacitor’s
potential difference.
Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a
parallel-plate capacitor is V  Es where s is the distance from the negative electron. The proton has charge q  e
and its potential energy at a point where the capacitor’s potential is V is U  eV. The proton will gain potential
energy
U  eV  e(250 V)  1.60  1019 C (250 V)  4.0  1017 J
if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic
energy which is
1
2
1.67  10
27

kg  200,000 m/s  3.34  10 17 J
2
This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach
the positive plate. Thus the proton does not reach the plate because K < U.
(b) The energy-conservation equation Kf  Uf  Ki  Ui is
1
2
 vf  vi2 
2q
Vi  Vf  
m
mvf2  qVf  21 mvi2  qVi  21 mvf2  21 mvi2  q Vi  Vf 


2
2.0  105 m/s 


2 1.60  10 19 C  250 V  0 V 
1.67  10 27 kg
2Q
b
Problem: Two identical 7.5 µC point charges are initially at rest and spaced 5.5 cm from
each other. If they are released at the same instant moment from rest, how fast will they
be moving when they are very far away from each other? Assume that they have identical
masses, 1.0 mg.
K  21 mv2 
+
 2.96  105 m/s
–
– 3Q