Question

Air breaks down and conducts charge as a spark if the electric field magnitude exceeds V/m. Determine the maximum charge that can be stored on an air-filled parallel-plate capacitor with a plate area of . A air-filled parallel-plate capacitor stores charge . Find the potential difference across its plates.

Answer 1

Answer:

Explanation:

Suppose

Magnitude of Electric Field is E V/m

Area of the cross-section is A

capacitor $C=\frac{\epsilon A}{d}$

Distance between Area of capacitor is d

Maximum Charge stored is

$Q_{max}=capacitor\times Potential\ Difference$

$Potential\ Difference=electric\ Field\times distance=E\times d$

$Q_{max}=C\times Ed=CEd$

$Q_{max}=\frac{\epsilon _0A}{d}\times Ed$

$Q_{max}=\epsilon _0AE$