Answer
(a) it is two times greater
Work Step by Step
The relation between electric field strength and the potential difference is
$E=Vd$
In the first case V=6.0 V and in the second case V=12.0 V.
d is the same in both cases.
Therefore, $E_{1}=6.0\,V\times d$ and
$E_{2}=12.0\,V\times d=2(6.0\,V\times d)=2E_{1}$.
In the second case, the electric field strength is two times greater.
