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uA parallel plate capacitor has a capacitance 50 μF and 100 μF when entered in a oil the electric condstant of the oil is?
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Capacitance = k ε₀A / d
When a dielectric is added between plates of capacitor then its Capacitance increases by k times (where k = dielectric constant)
Final Capacitance = k × Initial Capacitance
k = Final Capacitance / Initial Capacitance
= 100 μF / 50 μF
= 2
Dielectric constant of oil is 2
When a dielectric is added between plates of capacitor then its Capacitance increases by k times (where k = dielectric constant)
Final Capacitance = k × Initial Capacitance
k = Final Capacitance / Initial Capacitance
= 100 μF / 50 μF
= 2
Dielectric constant of oil is 2
Eva92:
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You’re welcome
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