Question

The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as `A,B,C…. .` The arrangements lies in air (Take `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)]`

A positive charge is placed at `B`,When it is released:
A. no force will be exerted on it .
B. it will move towards `A`
C. it will move towards `C`
D. it will move towards `E`

Answer 1

Correct Answer - B
Direction of `E.F` at `B`is towards `A` will exert froce in this direction only, causing the positive charge to move .[`vec(E)` is perpendicular to equipotential surface and its direction is form high potential to low potential].