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Question

(D) The electric field $E$ in the presence of a dielectric between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}

Vedclass · Accepted Answer

$6 	imes 10^{-7} \ Cm^{-2}$

Vedclass · Answer

(D) The electric field $E$ in the presence of a dielectric between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}{K \varepsilon_0}$where $\sigma$ is the surface charge density,$K$ is the dielectric constant,and $\varepsilon_0$ is the permittivity of free space.Rearranging the formula to solve for the charge density $\sigma$:$\sigma = K \varepsilon_0 E$Given values:$K = 2.2$$\varepsilon_0 \approx 8.85 	imes 10^{-12} \ F/m$$E = 3 	imes 10^4 \ V/m$Substituting these values into the equation:$\sigma = 2.2 	imes (8.85 	imes 10^{-12}) 	imes (3 	imes 10^4)$$\sigma = 2.2 	imes 8.85 	imes 3 	imes 10^{-8}$$\sigma \approx 58.41 	imes 10^{-8} \ C/m^2$$\sigma \approx 5.84 	imes 10^{-7} \ C/m^2$Rounding to the nearest value,we get $\sigma \approx 6 	imes 10^{-7} \ C/m^2$.

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

Similar Questions

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