Complete Question:
Two small objects each with a net charge of Q (where Q is a positive number) exert a force of magnitude "F" on each other. We
replace one of the objects with another whose net charge is 4Q. The original magnitude of the force on the Q charge was "F"; what is the magnitude of the force on the Q charge now?
Answer:
4 F₀
Explanation:
Assuming that we can treat to both objects as point charges, we can find the force "F" that one charge exerts upon the other applying Coulomb´s law, as follows:
F₀ = K*Q₀² / r₁₂²
If we replace one of the charges by one with a 4Q₀ charge, the new value of F will be as follows:
F₁ = K*Q₀*4Q₀ / r₁₂² =( K*Q₀² / r₁₂²)* 4 = 4* F₀
This value is reasonable, as the electrostatic force is a linear - type one, so it is possible to use the superposition principle (we can get the force exerted by one charge on another without considering the ones due to another charges)
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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
True, the wavelength dies down due to high frequency and low amptitude.