Question

An infinite number of charges each equal to q, are placed along the X-axis at `x = 1, x = 2, x = 4, x = 8`,…….. and so on.
(i) find the electric field at a point `x = 0` due to this set up of charges.
(ii) What will be the electric field if the above setup, the consecutive charges have opposite signs.

Answer 1

The electric field intenistites due to position charges and due to -ve charge the fields intensity is towards the charges
The resultant intensity at the origin
`E=E_(1)+E_(3)-E_(4).......`
`E=(Q)/(4pi epsilon_(0))(1-(1)/(2^(2))+(1)/(4^(2))-(1)/(8^(2))+....)`
since the expression in the brackets is in `GP` with a common ratio`=(-1)/(2^(2))=-(1)/(4)`
`E=(Q)/(4piepsilon_(0))(1)/([1-((-1)/(4))])=(Q)/(4piepsilon_(0))(4)/(5)`
` E=(4)/(5)(Q)/(4piepsilon_(0))`
`E=(Q)/(5piepsilon_(0))`