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Sagot :
To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find
[tex]d = \frac{ \:Δv}{e} [/tex]
[tex] = \frac{1.5v}{100v/m} [/tex]
[tex] = 1.5 \times 10 {}^{ - 2} m[/tex]
learn more about potential difference from here: https://brainly.com/question/28166044
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