Question

5.
«A parallel plate capacitor is charged by a supply of 500V. Plate separation is 1mm. If capacitor is
lowered into water with water filling in the gap between the plates, then change in pressure at any
point between the plates is kɛ, X10'3(N/m)2. Find the value of K, given €, (for water)=81.​

Answer 1

Explanation:

Given A parallel plate capacitor is charged by a supply of 500 V. Plate separation is 1 mm. If capacitor is  lowered into water with water filling in the gap between the plates, then change in pressure at any  point between the plates is kɛ, X10'3(N/m)2. Find the value of K, given €, (for water)=81.

• Now there are two parallel plate capacitors and is lowered into water and the gap is also filled with water.. There is positive charge on one plate and negative charge on another plate.
• So initial pressure Pi = 1/2 εo E^2 (where E is the electric field
•                                 = ½ εo (v / d)^2
• Now final pressure Pf = 1/2 εo ε^2 (v/d)^2 (since dielectric ) (voltage is constant and distance is same)
•             So Δ P = Pf – Pi
•                         = ½ εo ε^2 (v/d)^2 – ½ εo (v/d)^2
•                        = ½ εo ε (v/d)^2 (ε – 1)
•                                  = ½ x 8.854 x 10^-12 x 81 (500 / 1 x 10^-3)^2 (81 – 1)
•                                   = 4.427 x 10^-12 x 81 (500 x 500 / 1 x 10^-3 x 10^-3) x 80
•                                      = 7171.74
•                 Or ΔP = 7.17 K Pa

Reference link will be

https://brainly.in/question/17481246