Two parallel plates, each of area 7.50 cm^2, are separated by 2.50 mm of water.

Question

Two parallel plates, each of area 7.50 cm^2, are separated by 2.50 mm of water. A voltage of 7.00 V is applied across the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the charge stored on each plate? If the voltage across the plates remains 7.00 V when the water is replace by air what is t new charge stored on each plate? A parallel plate capacitor filled with paper has 2.00 middot 10^-6 J of potential gravitational energy [position (i)]. The distance between its plates is 2.50 mm and it has a plate area of 5.00 middot 10^-3 m^2. The paper is removed without losing any of the charge on the capacitor [position (ii)]. What is the voltage on the capacitor without the paper?

Explanation / Answer

(3)ans

Given that

area A=7.5 cm^2

distance b/w plates d=2.5 mm

voltage V=7v

now we find the electric field b/w plates

electric field E=V/d=7/2.5*10^-3=2.8*10^3 N/c

now we find the charge stored on each plate

capacitance C=keA/d=80*8.85*10^-12*7.5*10^-4/2.5*10^-3

=2.124*10-10 F

charge q=CV=2.124*10^-10*80=169.92*10^-10 C

---------------------------------------------------------------------------------------------------------------------------------------------------

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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