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



.