What is the electrostatic pressure on the plate of a PPC of capacitance 2 μ F separated by 2 mm and 20 V potential difference is applied across the capacitor

Question

What is the electrostatic pressure on the plate of a PPC of capacitance 2 μ F separated by 2 mm and 20 V potential difference is applied across the capacitor (A) 2.425 × 10-5 Pa (B) 4.425 × 10-4 Pa (C) 7.425 × 10-4 Pa (D) 1.425 × 10-5 Pa

Tardigrade Admin · Accepted Answer

C =2 μ F d =2 mm V =20 v P =( F / A )=(σ2/2 ε0)=(1/2) ∈ E 2 Where E =( V / d ) =(1/2) ∈0 ( v 2/ d 2) =(1/2) × 8.85 × 10-12 × (400/4 × 10-6) P =4.425 × 10-4 Pa

Q. What is the electrostatic pressure on the plate of a PPC of capacitance $2 μ F$ separated by $2 mm$ and $20 V$ potential difference is applied across the capacitor

$2.425 \times 1 0^{- 5} P a$

$4.425 \times 1 0^{- 4} P a$

$7.425 \times 1 0^{- 4} P a$

$1.425 \times 1 0^{- 5} P a$

$C = 2 μ F$
$d = 2 mm$
$V = 20 v$
$P = \frac{F}{A} = \frac{σ ^{2}}{2 ε _{0}} = \frac{1}{2} \in E^{2}$
Where $E = \frac{V}{d}$
$= \frac{1}{2} \in_{0} \frac{v ^{2}}{d ^{2}}$
$= \frac{1}{2} \times 8.85 \times 1 0^{- 12} \times \frac{400}{4 \times 1 0 ^{- 6}}$
$P = 4.425 \times 1 0^{- 4} P a$

Q. What is the electrostatic pressure on the plate of a PPC of capacitance 2μF separated by 2mm and 20V potential difference is applied across the capacitor

2.425×10−5Pa

4.425×10−4Pa

7.425×10−4Pa

1.425×10−5Pa