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(C) The electric field $E$ between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}{\varepsilon_0} = \frac{q}{A \va

Vedclass · Accepted Answer

$8.85 	imes 10^{-10}\,C$

Vedclass · Answer

(C) The electric field $E$ between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}{\varepsilon_0} = \frac{q}{A \varepsilon_0}$Where $q$ is the charge on the plate,$A$ is the area of the plate,and $\varepsilon_0$ is the permittivity of free space $(8.85 	imes 10^{-12}\,F/m)$.Rearranging the formula to solve for $q$:$q = E \cdot A \cdot \varepsilon_0$Substituting the given values ($E = 100\,N/C$,$A = 1\,m^2$,$\varepsilon_0 = 8.85 	imes 10^{-12}\,F/m$):$q = 100 	imes 1 	imes 8.85 	imes 10^{-12}$$q = 8.85 	imes 10^{-10}\,C$Therefore,the magnitude of the charge on each plate is $8.85 	imes 10^{-10}\,C$.

$A$ parallel plate capacitor with plates of area $1\,m^2$ each,are at a separation of $0.1\,m.$ If the electric field between the plates is $100\,N/C,$ what is the magnitude of the charge on each plate?

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