In a thundercloud there may be an electric charge of 41 C near the top of the cloud and −41 C near the bottom of the cloud. If t
hese charges are separated by about 8 km, what is the magnitude of the electric force between these two sets of charges? The value of the electric force constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of N.
1 answer:
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Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F =
q =
we calculate
q =
q = Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by
Substitute the values then we get
Capacitance between parallel plate capacitor=
V=
Hence, the charge resides on the outer surface=