What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ?

Question

What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ? (A) 1.694×109m2 (B) 4.529×109m2 (C) 9.281×109m2 (D) 12.981×109m2

Tardigrade Admin · Accepted Answer

The expression of parallel plate capacitor is given by C =( A ϵ0/ d ) so area of plate, A=(C d/ϵ0)=(3 ×(5 × 10-3)/8.85 × 10-12) =1.694 × 109 m 2

Q. What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ?

$1.694 \times 1 0^{9} m^{2}$

$4.529 \times 1 0^{9} m^{2}$

$9.281 \times 1 0^{9} m^{2}$

$12.981 \times 1 0^{9} m^{2}$

The expression of parallel plate capacitor is given by
$C = \frac{A ϵ _{0}}{d}$ so area of plate,
$A = \frac{C d}{ϵ _{0}} = \frac{3 \times ( 5 \times 1 0 ^{- 3} )}{8.85 \times 1 0 ^{- 12}}$
$= 1.694 \times 1 0^{9} m^{2}$

Q. What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm ?

1.694×109m2

4.529×109m2

9.281×109m2

12.981×109m2

The expression of parallel plate capacitor is given by C=dAϵ0​​ so area of plate, A=ϵ0​Cd​=8.85×10−123×(5×10−3)​ =1.694×109m2

$1.694 \times 1 0^{9} m^{2}$

$4.529 \times 1 0^{9} m^{2}$

$9.281 \times 1 0^{9} m^{2}$

$12.981 \times 1 0^{9} m^{2}$

The expression of parallel plate capacitor is given by
$C = \frac{A ϵ _{0}}{d}$ so area of plate,
$A = \frac{C d}{ϵ _{0}} = \frac{3 \times ( 5 \times 1 0 ^{- 3} )}{8.85 \times 1 0 ^{- 12}}$
$= 1.694 \times 1 0^{9} m^{2}$