Answer:
No number answer; don't want to pull out a calculator lol
Explanation:
Capacitors are added in parallel; opposite of resistors.
So for 9, add 10 and 2.5 then do ((1/12.5)+(1/.3))^-1.
For 10, add 0.75 and 15 first, then the rest is the same idea as 9.
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
Explanation:
PEgrav = m *• g • h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.
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Potential Energy - The
Answer:
The maximum value of the induced magnetic field is .
Explanation:
Given that,
Radius of plate = 30 mm
Separation = 5.0 mm
Frequency = 60 Hz
Suppose the maximum potential difference is 100 V and r= 130 mm.
We need to calculate the angular frequency
Using formula of angular frequency
Put the value into the formula
When r>R, the magnetic field is inversely proportional to the r.
We need to calculate the maximum value of the induced magnetic field that occurs at r = R
Using formula of magnetic filed
Where, R = radius of plate
d = plate separation
V = voltage
Put the value into the formula
Hence, The maximum value of the induced magnetic field is .