29. Consider an infinitesimal segment of the rod, located between x and x+dx. It has length dx and contains
charge dq = λ dx = cx dx. Its distance from P
1
is d + x and the potential it creates at P
1
is
dV =
1
4πε
0
dq
d + x
=
1
4πε
0
cx dx
d + x
.
To find the total potential at P
1
, integrate over the rod:
V =
c
4πε
0
L
0
x dx
d + x
=
c
4πε
0
[x
− d ln(x + d)]
L
0
=
c
4πε
0
L
− d ln
1 +
L
d
.