Answer
2) 1.5×10-2 m
Explanation
The potential difference is related to the electric field by:
(1)
where
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
![E=1 V/cm=100 V/m](https://tex.z-dn.net/?f=E%3D1%20V%2Fcm%3D100%20V%2Fm)
when connected to a battery with potential difference
![\Delta V=1.5 V](https://tex.z-dn.net/?f=%5CDelta%20V%3D1.5%20V)
Solving the equation (1) for d, we find
![d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%5CDelta%20V%7D%7BE%7D%3D%5Cfrac%7B1.5%20V%7D%7B100%20V%2Fm%7D%3D0.015%20m%3D1.5%20%5Ccdot%2010%5E%7B-2%7D%20m)
Sexual reproduction. thats the answer i think
Answer:
Oooo someone is writing a answer. (Also im new to this so Idk what to really do.)
Explanation: