A charge + 1 6 C is fixed at each of the points x = 3 , 9 , 1 5 , … , ∞ on the x -axis, and a charge − 1 6 C is fixed at each of the points x = 6 , 1 2 , 1 8 , … , ∞ on the x -axis. Then find the potential at the origin due to these charges.
If the potential is of the form C π ε 0 A ln B , find A + B + C , where A and C are co-prime and B is prime.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Shouldn't you be telling why that series is ln(2)?
Log in to reply
It is an expansion of ln(2). I learnt it from Wikipedia
Hint: Use the fact that harmonic series can be expressed as
H ( n ) = l n ( n ) + γ + α 0
Where γ is some constant and α 0 is some zero-sequence.
This series the expansion of ln(1+x).Here the value of x is 1.
Good Question!! Solved it the same way. But check your solution. A kiddish error. 4+2+3 = 9. :D
Log in to reply
Thanks for the report it was really kiddish :P
However,this series is conditionally convergent,and it's terms can't be reordered.So can you explain why you have used this particular arrangement:see riemann's rearrangement theorem.
It is proportional to inverse of distance squared. So, where has the square gone. Please let me know
I am in second grade P
Problem Loading...
Note Loading...
Set Loading...
Potential energy at the origin is given by:
V = 4 π ε 0 1 6 ( 3 1 − 6 1 + 9 1 − 1 2 1 + . . . ) = 4 π ε 0 1 6 . 3 1 ( 1 1 − 2 1 + 3 1 − 4 1 + . . . ) = 3 π ε 0 4 . l n ( 2 ) ∴ A + B + C = 4 + 2 + 3 = 9