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(D) The electric field $E$ inside a dielectric-filled capacitor is given by $E = \frac{Q}{K A \varepsilon_0}$,where $Q$ is the charge,$K$ is the diele

Vedclass · Accepted Answer

$8.4 	imes 10^{-6} \ C$

Vedclass · Answer

(D) The electric field $E$ inside a dielectric-filled capacitor is given by $E = \frac{Q}{K A \varepsilon_0}$,where $Q$ is the charge,$K$ is the dielectric constant,$A$ is the area,and $\varepsilon_0$ is the permittivity of free space.To avoid dielectric breakdown,the electric field must not exceed the dielectric strength $E_{max} = 1.9 	imes 10^7 \ V/m$.Thus,$Q_{max} = E_{max} \cdot K \cdot A \cdot \varepsilon_0$.Given: $A = 100 \ cm^2 = 10^{-2} \ m^2$,$K = 5.0$,$\varepsilon_0 = 8.85 	imes 10^{-12} \ F/m$,and $E_{max} = 1.9 	imes 10^7 \ V/m$.Substituting the values:$Q_{max} = (1.9 	imes 10^7) 	imes 5.0 	imes 10^{-2} 	imes 8.85 	imes 10^{-12}$$Q_{max} = 8.4075 	imes 10^{-6} \ C \approx 8.4 	imes 10^{-6} \ C$.

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