A water-filled parallel plate capacitor has a plate area of 2.9 cm2 and plate se

Question

A water-filled parallel plate capacitor has a plate area of 2.9 cm2 and plate separation of 1.7 mm. The potential difference between its plates is held at 5.4 V. Calculate the magnitude of the electric field between its plates, the charge stored on each plate, and the charge stored on each plate after water is replaced by air (a) the magnitude of the electric field between its plates 3176 (b) the charge stored on each plate 0.66 (c) the charge stored on each plate after water is replaced by air 8.15 PC

Explanation / Answer

Dielectric constant of water(K) is 80.

So answer what I am getting is 0.6522 nC

When air is in between then charge is q = CV

When water is in between then the charge will be q' = KCV.

So q' = 80×8.15×10-12 = 652 pC = 0.652 nC.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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