Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric fi
eld E between the plates? Express E in terms of q and other quantities given in the introduction, in addition to ϵ0 and any other constants needed..
1 answer:
Let A represent the area of each plate. According to Gauss's Law, the electric field present between the plates can be derived.
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Answer:
Explanation:
Provided:
fiber diameter d= 18 μm
screen distance D= 30 cm
wavelength λ= 560 nm
from this, we can determine the fringe width
substituting the values yield
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09
Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489
Thus, the acceleration is approximately 5.5 ft/s²!!
