IIT JEE Advanced 2012 Paper 2

It can be seen, from the figure depicted here, that
|

→

E

34| = |

→

E

12| = 2K.
Therefore,

→

E

34+

→

E

12 =

→

E

′ = 2K is along OD.
Also the net field

→

E

=

→

E

′+

→

E

56 = 2K + 4K = 6K is along OD.

The three sets of the dipoles create a zero potential at the centre.
The vertical line PR is along the equatorial axis of three sets of dipoles.
Now, the potential at all points along the line PR is zero.
The line ST lies along the axis of the dipole (+2q, –2q) and the potential varies from point to point.