A parallel-plate air capacitor has a capacitance of 500 pF and a charge of magni

Question

A parallel-plate air capacitor has a capacitance of 500 pF and a charge of magnitude 0.200 uC on each plate. The plates are 0.700 mm apart.


Part A
What is the potential difference between the plates?


Part B
What is the area of each plate?


Part C
What is the electric-field magnitude between the plates?


Part D
What is the surface charge density on each plate?

Explanation / Answer

A)Q=CV =>V = Q/C = (0.2*(10^(-6)))/(500*(10^(-12))) = 400V B)C = 500*(10^(-12)) =>(8.85*(10^(-12)))*A/(0.7*(10^(-3))) = 500*(10^(-12)) =>A = 0.03955m^2 D)Sigma = Q/A = 7.5857*(10^(-6)) = Surface charge density C)Electric field = Sigma/Epsilon = 857142.8571

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.