B, the surface of a flat table.
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance

∴ Area of each plate is 7.9060 m²
Voltage

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
I think there is only one.
<span>
</span>Both are conserved<span>.</span>
Evaporation (or another word to use is water vapor.)