A large parallel plate capacitor, whose plates have an area of 1 m2 and are separated from each other by 1 mm , is being charged at a rate of 25 V s- 1 . If the dielectric between the plates has the dielectric constant 10 , then the displacement current at this instant is

Question

A large parallel plate capacitor, whose plates have an area of 1 m2 and are separated from each other by 1 mm , is being charged at a rate of 25 V s- 1 . If the dielectric between the plates has the dielectric constant 10 , then the displacement current at this instant is (A) 25 μ A (B) 11 μ A (C) 2.2 μ A (D) 1.1 μ A

Tardigrade Admin · Accepted Answer

C=((ϵ )0 K A/d)=((. 8.85 × (10)- 12 .) × 10 × 1/(10)- 3) =8.85× 10- 8 F i=(d/d t)(C V)=C(d V/d t)=8.85× (10)- 8× 25 =2.2× 10- 6 A

Q. A large parallel plate capacitor, whose plates have an area of $1 m^{2}$ and are separated from each other by $1 mm$ , is being charged at a rate of $25 V s^{- 1}$ . If the dielectric between the plates has the dielectric constant $10$ , then the displacement current at this instant is

$25 μ A$

$11 μ A$

$2.2 μ A$

$1.1 μ A$

$C = \frac{( ϵ ) _{0} K A}{d} = \frac{( 8.85 \times ( 10 ) ^{- 12} ) \times 10 \times 1}{( 10 ) ^{- 3}}$
$= 8.85 \times 1 0^{- 8} F$
$i = \frac{d}{d t} (C V) = C \frac{d V}{d t} = 8.85 \times (10)^{- 8} \times 25$
$= 2.2 \times 1 0^{- 6}$ A

Q. A large parallel plate capacitor, whose plates have an area of 1m2 and are separated from each other by 1mm , is being charged at a rate of 25Vs−1 . If the dielectric between the plates has the dielectric constant 10 , then the displacement current at this instant is

25μA

11μA

2.2μA

1.1μA