Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $\mathbf{E}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$. When the space is filled with dielectric, the electric field is$\mathbf{E}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$ E = 2.50 * 105 V>m. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Short answer

(a) The charge density on each surface of the dielectric is $0.620\mathrm{\mu C}/{\mathrm{m}}^2$

(b) The dielectric constant is 1.28

Step-by-step solution

The formula of the charge density on the surface and plates and dielectric constant

${\mathbf{\sigma }}_{\mathbf{I}}\mathbf{=}\mathbf{\sigma }\mathbf{(}\mathbf{1}\mathbf{‐}\frac{\mathbf{1}}{\mathbf{K}}\mathbf{)}\mathbf{}\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

where${\sigma }_{\mathrm{I}}$ is the charge density on the dielectric material,role="math" localid="1664257274567" $\sigma$ 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

$\mathbf{\sigma }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{o}}{\mathit{E}}_{\mathbf{o}}\mathbf{}\_\_\_\_\_\_\_\_\_\_\_\_\left(3\right)$

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

Calculate the surface charge density on the dielectric

(a)

We are given that

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

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

using equation (2)

the electric field between the plates

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

using equation (3)

the charge densityon each surface of the dielectric

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

using the equation (1)

the charge densityon the dielectric material

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

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

Calculate the dielectric constant

(b)

using equation (2)

the dielectric constant

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

Hence, the dielectric constant is 1.28