Chapter 15
Problem 1
F =
kqq
r2
Plugging in the numbers given, one finds F = −1.1 · 109 N where the negative sign means the charges
attract each other.
Problem 13
Consider the components of forces from each of the two charges. For the x direction, we need the cosine.
The y direction uses the sine.
9
−6
9
−6
P
·4·10−6
·2·10−6
cos 60 + (9·10 )7·10
cos 60 = 0.756N to the right
Fx = (9·10 )7·10
0.52
0.52
P
(9·109 )7·10−6 ·2·10−6
(9·109 )7·10−6 ·4·10−6
Fy =
sin 60 −
sin 60 = 0.436N downwards
0.52
0.52
The y direction forces are subtracted because they are in opposite directions. The x direction forces are
F
in the same direction, so they add. The angle is given by θ = tan−1 Fxy = 30◦ . The magnitude of F is given
by the pythagorean theorem. The answer is:
F = 0.873N @30 degrees down from the x axis
Problem 15
The charges are in equilibrium, meaning the net force is zero. The three forces are gravity (y direction),
coulomb force (x direction) and tension along the string. Adding components for the charge on the right:
P
kq 2
P Fx = r2 − T sin 5 = 0
Fy = −mg + T cos 5 = 0
Solve for T in the Fy equation, plug it in for T in the Fx equation. The last thing is to find r, the distance
between the charges. Looking at the diagram, this distance is r = 2(0.3m) sin 5. Plug in all the numbers and
solve for q to find
|q| = 7.2nC
Problem 17
The electric force
P is given by F = qE, and is directed upwards, opposing gravity. Adding the forces in the y
direction gives
F = qE − mg = 0. This gives mg = qE which one can plug in numbers to find
m = 1.5 · 10−3 kg
Problem 23
−27
The force again is given by F = qE = ma. Therefore a = qE
kg, its
m . The mass of the proton is 1.67 · 10
charge is the same of that of an electron. The time taken to reach this speed is given by the definition of
1 2
1
2
acceleration, a = ∆v
∆t . The distance travelled by x = 2 at , and the kinetic energy by KE = 2 mv .
a = 6.13 · 1010 m/s2 , ∆t = 1.96 · 10−5 sec, δx = 11.74m, and KE = 1.20 · 10−15 J
Problem 27
The total field is the sum of the field from the two charges. Let the left charge be at x = 0, and the right
charge at x = 1m. We add the two up and set them equal to zero, then solve for x.
P
9
−6
9
·6·10−6
E = 0 = − 9·10 ·2.5·10
+ 9·10(x−1)
2
x2
Rearrange the equation and cancel some stuff to get 2.5(x − 1)2 = 6x2 . Expand the binomial and solve
for x with the quadratic equation. The two solutions are x = −1.82m, x = 0.39m. The only one that works
1
is the x = −1.82m one, since the other solution is in between both charges (in this region, the charge would
not be in equilibrium, both of the other charges push to the left). Thus,
x = −1.82m, or 1.82m left of the -2.5µC charge.
Problem 43
Gauss’ Law says that the electric field is proportional to the total charge enclosed. If we are outside the
spherical shell, a Gaussian surface would enclose a positive charge q, and a negative charge q, adding up to
zero. When inside the shell, a Gaussian surface only encloses the positive charge at the center, thus
The field outside the shell is zero, the field inside is that due to a point charge, namely E =
2
kq
r2