Question

A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate? What is the potential difference between the plates if the capacitor is charged to 1.2 mu C? What is the stored energy? How much charge can the capacitor carry before a spark occurs between the two plates (thus discharging the capacitor)?

Answer 1

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Step-by-step 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

`C=(\epsilon_(0)A)/(d)\\\Rightarrow A=(Cd)/(\epsilon_(0))\\\Rightarrow A=(0.14* 10^(-6)* 0.5* 10^(-3))/(8.854* 10^(-12))\\\Rightarrow A=7.9060\ m^2`

∴ Area of each plate is 7.9060 m²

Voltage

`V=(Q)/(C)\\\Rightarrow V=(1.2* 10^(-6))/(0.14* 10^(-6))\\\Rightarrow V=8.57\ Volts`

∴ 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