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A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ = 7.4 × 10–12Ω–1m–1. If the charge on the capacitor at the instant t = 0 is q0 = 8.55µC, then calculate the leakage current (in µA) at the instant t = 12s.
Correct answer is '0.193'. Can you explain this answer?
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A leaky parallel plate capacitor is filled completely with a material ...
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The correct answer is: 0.193
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Now current as function of time
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The correct answer is: 0.193
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A leaky parallel plate capacitor is filled completely with a material ...
Σ=2x10^-6Ω^-1m^-1. The distance between the plates is 0.02m and the area of each plate is 0.01m^2.
The leakage current through the capacitor can be calculated using the formula:
I = K * σ * A / d
Where:
I = leakage current
K = dielectric constant (5)
σ = electrical conductivity (2x10^-6 Ω^-1m^-1)
A = area of each plate (0.01m^2)
d = distance between the plates (0.02m)
Plugging in the values:
I = 5 * 2x10^-6 * 0.01 / 0.02
I = 5 * 2x10^-8 / 0.02
I = 10x10^-8 / 0.02
I = 0.5x10^-6
I = 5x10^-7 A
Therefore, the leakage current through the capacitor is 5x10^-7 Amperes.
The leakage current through the capacitor can be calculated using the formula:
I = K * σ * A / d
Where:
I = leakage current
K = dielectric constant (5)
σ = electrical conductivity (2x10^-6 Ω^-1m^-1)
A = area of each plate (0.01m^2)
d = distance between the plates (0.02m)
Plugging in the values:
I = 5 * 2x10^-6 * 0.01 / 0.02
I = 5 * 2x10^-8 / 0.02
I = 10x10^-8 / 0.02
I = 0.5x10^-6
I = 5x10^-7 A
Therefore, the leakage current through the capacitor is 5x10^-7 Amperes.
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A leaky parallel plate capacitor is filled completely with a material having dielectric constantK=5and electrical conductivityσ = 7.4×10–12Ω–1m–1. If the charge on the capacitor at the instantt= 0isq0= 8.55µC, then calculate the leakage current (inµA) at the instantt= 12s.Correct answer is '0.193'. Can you explain this answer?
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