91影视

${\mathbf{蟽}}_{\mathbf{I}}\mathbf{=}\mathbf{蟽}\mathbf{(}\mathbf{1}\mathbf{鈥�}\frac{\mathbf{1}}{\mathbf{K}}\mathbf{)}\mathbf{}\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

where${蟽}_{\mathrm{I}}$ is the charge density on the dielectric material,role="math" localid="1664257274567" $蟽$ is the surface charge density and K is the factor by which potential difference between the plates decreases.

$\mathbf{K}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{o}}}{\mathbf{E}}\mathbf{}\mathrm{\_}\_\_\_\_\_\_\_\_\_\_\_\left(2\right)$

whererole="math" localid="1664257303665" ${\mathrm{E}}_{\mathrm{o}}$ is the electric field of the vacuum and E is the electric field of the dielectric

$\mathbf{蟽}\mathbf{=}{\mathit{蔚}}_{\mathbf{o}}{\mathit{E}}_{\mathbf{o}}\mathbf{}\_\_\_\_\_\_\_\_\_\_\_\_\left(3\right)$

where${蔚}_o$ is the universal constant equal to $8.854脳{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$

(a)

We are given that

the electric field between the two plates ${\mathrm{E}}_{\mathrm{o}}=3.20脳{10}^5\mathrm{V}/\mathrm{m}$

the electric field with a dielectric material is $\mathrm{E}=2.50脳{10}^5\mathrm{V}/\mathrm{m}$

using equation (2)

the electric field between the plates

$$\begin{gathered} \mathrm{K}=\frac{{\mathrm{E}}_{\mathrm{o}}}{\mathrm{E}} \\ =\frac{3.20脳{10}^5\mathrm{V}/\mathrm{m}}{2.50脳{10}^5\mathrm{V}/\mathrm{m}} \\ =1.28 \end{gathered}$$

using equation (3)

the charge densityon each surface of the dielectric

$$\begin{gathered} \mathrm{蟽}={\mathrm{蔚}}_{\mathrm{o}}{\mathrm{E}}_{\mathrm{o}} \\ =\left(8.854脳{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)\left(3.20脳{10}^5\mathrm{V}/\mathrm{m}\right) \\ =2.833脳{10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$

using the equation (1)

the charge densityon the dielectric material

$$\begin{gathered} {\mathrm{蟽}}_{\mathrm{I}}=\mathrm{蟽}\left(1-\frac{1}{\mathrm{K}}\right) \\ =\left(2.833脳{10}^{-6}\mathrm{C}/{\mathrm{m}}^2\right)\left(1-\frac{1}{1.28}\right) \\ =0.620脳{10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ =0.620\mathrm{渭颁}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the charge density on each surface of the dielectric is $0.620\mathrm{渭颁}/{\mathrm{m}}^2$

(b)

using equation (2)

the dielectric constant

$$\begin{gathered} \mathrm{K}=\frac{{\mathrm{E}}_{\mathrm{o}}}{\mathrm{E}} \\ =\frac{3.20脳{10}^5\mathrm{V}/\mathrm{m}}{2.50脳{10}^5\mathrm{V}/\mathrm{m}} \\ =1.28 \end{gathered}$$

Hence, the dielectric constant is 1.28