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



learn more about potential difference from here: brainly.com/question/28166044
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No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s so you are left with Solution 4.78 10-19 J OR .478 aJ <span>Apex - 467 nm ^.^ hopefully thats the correct thing</span>
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