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



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The density of ice is less than the density of water. C
The difference in electric potential energy between the two points is

where q is the magnitude of the charge and

is the electric potential difference.
But for energy conservation, the difference in electric potential energy

between the two points is equal to the work done to move the charge between A and B:

so we have

and by substituting the numbers of the problem, we find the value of

: