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A charge +q is fixed at each of the points x = x. , x = 3x. , x = 5x. and so on to infinity on the x-axis and a charge -q is fixed at each of the points x = 2x. , x = 4x. , x = 6x. and so on to infinity. Here x. is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4*pie*epsilon.*r). Then the potential at the origin to the above system is:-

A charge +q is fixed at each of the points x = x. , x = 3x. , x = 5x. and so on to infinity on the x-axis and a charge -q is fixed at each of the points x = 2x. , x = 4x. , x = 6x. and so on to infinity. Here x. is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/(4*pie*epsilon.*r). Then the potential at the origin to the above system is:-

Grade:12th pass

1 Answers

Samyak Jain
333 Points
2 years ago
Electric potentials at origin due to charge +q fixed at x = x.,x = 3x., x = 5x.,... are Kq/x., Kq/3x.,Kq/5x.,...
Electric potentials at origin due to charge –q fixed at x = 2x.,x = 4x., x = 6x.,... are – Kq/2x., –Kq/4x., –Kq/6x.,...
Here K = 1/4\pi \varepsilon
\therefore Net electric potetial at origin is Kq/x. – Kq/2x + Kq/3x – Kq/4x + ….
   =  (Kq/x) (1 – 1/2 + 1/3 – 1/4 + …...)                           ….(1)
Now, ln(1 + x) = x – x2/2 + x3/3 + x4/4 – …...
Put x = 1.  \Rightarrow  ln(1 + 1) = 1 – 12/2 + 13/3 + 14/4 – …...
or  1 – 1/2 + 1/3 + 1/4 – …...  =  ln(2).   Substitute in (1).
\therefore Net electric potetial at origin is (Kq/x) ln2  or  (q ln2) / 4\pi \varepsilonx.

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