Question 654: Electrostats & Capacitance

Question

The capacitance of a parallel plate capacitor with air as medium is 6$$\mu $$F. With the introduction of a dielectric medium, the capacitance become 30 $$\mu $$F The permittivity of the medium is :$$\left( {{\varepsilon _0} = 8.85 	imes {{10}^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}} ight)$$

Accepted Answer

The capacitance increases with dielectric constant K
$$C = K{C_0}$$
(or)$$K = {C \over {{C_0}}} = {{30} \over 6} = 5$$
$$\varepsilon  = K{\varepsilon _0} = $$$$5 	imes 8.85 	imes {10^{ - 12}} =$$
$$ 0.44 	imes {10^{ - 10}}$$

Answer

$$1.77 	imes {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$$

Answer

$$0.44 	imes {10^{ - 10}}{C^2}{N^{ - 1}}{m^{ - 2}}$$

Answer

$$5.00{C^2}{N^{ - 1}}{m^{ - 2}}$$

Answer

$$0.44 	imes {10^{ - 13}}{C^2}{N^{ - 1}}{m^{ - 2}}$$

Electrostats & Capacitance

The capacitance of a parallel plate capacitor with air as medium is 6$\mu F. With the introduction of a dielectric medium, the capacitance become 30 \mu F The permittivity of the medium is :\left( {{\varepsilon _0} = 8.85 \times {{10}^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}} \right)

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