A parallel plate capacitor separated by an air gap is made of 0.8m2 tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each...
Question 1 Report
A parallel plate capacitor separated by an air gap is made of 0.8m2 tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each plate?
Take \( \epsilon_o = 8.85 * 10^{-12}\,\mathrm{F\,m^{-1}} \)
Answer Details
To calculate the charge on each plate of a parallel plate capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage applied.
The capacitance of a parallel plate capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the medium (in this case, air), A is the area of each plate, and d is the distance between the plates.
Given:
Area of each plate (A) = 0.8 m^2
Distance between the plates (d) = 20 mm = 0.02 m
Permittivity of air (ε) = 8.85 x 10^-12 F/m
Using the formula for capacitance, we can calculate C:
C = εA/d
= (8.85 x 10^-12 F/m)(0.8 m^2)/(0.02 m)
= 8.85 x 10^-12 F/m * 40 F
= 3.54 x 10^-10 F
Now, we can use the formula Q = CV to calculate the charge on each plate:
Q = (3.54 x 10^-10 F)(120 V)
= 4.25 x 10^-8 C
= 42.5 x 10^-9 C
= 42.5 nC
Therefore, the charge on each plate of the parallel plate capacitor is **42.5 nC**.
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