A force of 120 N is exerted on a 40 kg container which sits on a floor. If the frictional force between floor and container is 8
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Answer: 39.8 μC
Explanation:
The magnitude of the electric field generated by a capacitor is given by:
d is the distance between the plates.
For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.
where A is the area of the plate and ε₀ is the absolute permittivity.
substituting, we get
It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.
radius of the plates of the capacitor, r = 69 cm = 0.69 m
Area of the plates, A = πr² = 1.5 m²
Thus, the maximum charge that can be placed on disks without a spark is:
Q = E×ε₀×A
⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.