Question

A parallel-plate capacitor in air has a plate separation of 1.51 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 260 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) What is the charge on the plates before immersion?
(b) What is the charge on the plates after immersion?

Answer 1

Answer:

(a) 380.96 pC

(b) 3.25V

Explanation:

(a) Before immersion,

$C_{air}$ = $\frac{E_{0}A }{d}$

⇒(8.85E-12× 25E-4× 260) ÷(0.0151)

= 380.96 pC

(b)   Charge on the plates after immersion can be calculated by,

  Q = ΔV×C

      = Δ$V_{air}$ ÷ K

where K is the constant for distilled water

      = 260 ÷ 80

      = 3.25V

Answer 2

Answer:

Explanation:

(a) Charge  on the plates before immersion

as we know that,

C = εA/d

C = 8.85×$10^{-12}$× 25×$10^{-4}$/ 1.51×$10^{-2}$

   = 1.46×$10^{-12}$

Q = CV

   = (1.46×$10^{-12}$) (260)

   = 3.796×$10^{-10}$ C

(b) Charge on the plates after immersion

Q = 3.796×$10^{-10}$ C

The charge will remain the same, as the capacitor was disconnected before it was immersed.