Introductory Physics Volume Two

2.8 Homework 53
add up potential due to each small section making up the entire line:
∫
x+L ∫ Q
V (x) = dV ′ 1
L
=
dx′
4πɛ 0 x ′
x
= Q ( ) x + L
4πɛ 0 L ln x
Let’s check that this gives the expected result for L → 0, which will
result in Q being a point charge. We need the limit:
( ) (
x + L
lim ln = lim ln 1 + L )
≈ L
L→0 x L→0 x x
So
lim V (x) = Q
L→0 4πɛ 0 L · L
x = Q
4πɛ 0 x ,
which is indeed the electric potential a distance x from a point charge
Q.
§ 2.8 Homework
⊲ Problem 2.16
Through what potential difference would an electron need to be accelerated
for it to achieve a speed of 40% of the speed of light, starting
from rest?
⊲ Problem 2.17
How much work is required to move one mole of electrons from a region
where the electric potential is 9V to a region where the electric potential
is -5V?
⊲ Problem 2.18
An electron moving along the x axis has an initial speed of 2.7×10 6 m s
at
the origin. Its speed is reduced to 1.4 × 10 5 m s
at the point x = 2.0cm.
Calculate the potential difference between the origin and this point.
Which point is at the higher potential?
⊲ Problem 2.19
In Rutherford’s experiments alpha particles (charge +2e, mass 6.6 ×
10 −27 kg) were fired at a gold nucleus (charge +79e). An alpha particle
initially very far from the gold nucleus is fired at 2.0 × 10 7 m s
directly
toward the center of the nucleus. How close does the alpha particle get
to this center before turning around?
⊲ Problem 2.20
Show that the amount of work required to assemble four identical
charges Q at the corners of a square of side s is 5.41kQ 2 /s.