Answer:
W
Explanation:
= Temperature of the room = 22.0 °C = 22 + 273 = 295 K
= Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
= Surface area = 1.50 m²
= emissivity = 0.97
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as
W
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:
The potential difference between the plates is 596.2 volts.
Explanation:
Given that,
Capacitance
Charge
Separation of plates = 0.313 mm
We need to calculate the potential difference between the plates
Using formula of potential difference
Where, Q = charge
C = capacitance
Put the value into the formula
Hence,The potential difference between the plates is 596.2 volts.