A parallel plate capacitor with plates of area 1 m2 each, area t a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge each plate is: Take (ϵ0 = 8.85 × 10-12 (C2/N-m2))

Question

A parallel plate capacitor with plates of area 1 m2 each, area t a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge each plate is: Take (ϵ0 = 8.85 × 10-12 (C2/N-m2)) (A) 7.85 × 10-10 C (B) 6.85 × 10-10 C (C) 9.85 × 10-10 C (D) 8.85 × 10-10 C

Tardigrade Admin · Accepted Answer

E =(σ/ϵ0)=( Q / A ∈0) Q = AE ∈0 Q =(1)(100)(8.85 × 10-12) Q =8.85 × 10-10 C

Q. A parallel plate capacitor with plates of area $1 m^{2}$ each, area t a separation of $0.1 m$ . If the electric field between the plates is $100 N / C$ , the magnitude of charge each plate is :
Take $(ϵ_{0} = 8.85 \times 1 0^{- 12} \frac{C ^{2}}{N - m ^{2}})$

$7.85 \times 1 0^{- 10} C$

$6.85 \times 1 0^{- 10} C$

$9.85 \times 1 0^{- 10} C$

$8.85 \times 1 0^{- 10} C$

$E = \frac{σ}{ϵ _{0}} = \frac{Q}{A \in _{0}}$
$Q = A E \in_{0}$
$Q = (1) (100) (8.85 \times 1 0^{- 12})$
$Q = 8.85 \times 1 0^{- 10} C$

Q. A parallel plate capacitor with plates of area 1m2 each, area t a separation of 0.1m. If the electric field between the plates is 100N/C, the magnitude of charge each plate is : Take (ϵ0​=8.85×10−12N−m2C2​)

7.85×10−10C

6.85×10−10C

9.85×10−10C

8.85×10−10C