Mastering Physics: Exercise 24.3

Mastering Physics: Exercise 24.3

A parallel-plate air capacitor with a capacitance of 246 pF has a charge of magnitude 0.146 μC on each plate. The plates have a separation of 0.316 mm .

Part A

Question: What is the potential difference between the plates?

Answer: 

C = Q / V

=> V = Q / C

V = 0.146 ×10^(−6) / (246 ×10^(−12) )

V =  593  V  

Part B

Question: What is the area of each plate?

Answer: 

C = ϵ_0 × A / d

=> A = C × d / ϵ_0

A =  246 ×10^(−12)  × 0.316 ×10^(−3) /  ( 8.85 ×10^(−12) )

A =  8.78×10^(−3)  m^2  

Part C

Question: What is the electric field magnitude between the plates?

Answer: 

E = V / d

E = 593.4959 / (0.316 ×10^(−3) )

E =  1.88×10^6  V/m

Part D

Question: What is the surface-charge density on each plate?

Answer: 
σ = Q / A
σ =  0.146 ×10^(−6) / ( 8.78 ×10^(−3))
σ =  1.66×10^(−5)  C/m^2