Q .     The potential at a point `x` (measured in µm) due to some charges situated on the `x`-axis is given by `V(x)=20/(x^2-4)V`. The electric field `E` at `x=4mum` is given by

JEE 2007
A

`5/3` V/(mum) and in the -ve `x` direction

B

`5/3` V/mum and in the +ve` x` direction

C

`10/9` V/mum and in the -ve `x` direction

D

`10/9` V/mum and in the `+ve` `x` direction

HINT

`E=-((dV)/dx)`

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