Answer
2) 1.5×10-2 m
Explanation
The potential difference is related to the electric field by:
[tex]\Delta V=Ed[/tex] (1)
where
[tex]\Delta V[/tex] is the potential difference
E is the electric field
d is the distance
We want to know the distance the detectors have to be placed in order to achieve an electric field of
[tex]E=1 V/cm=100 V/m[/tex]
when connected to a battery with potential difference
[tex]\Delta V=1.5 V[/tex]
Solving the equation (1) for d, we find
[tex]d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m[/tex]