A parallel-plate capacitor has capacitance ${\mathbf{C}}_{\mathbf{O}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{pF}$ when there is air between the plates. The separation between the plates is $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{mm}$. (a) What is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$? (b) A dielectric with$\mathbf{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{70}$ is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$?

${\mathbf{Q}}_{\mathbf{o}}\mathbf{=}{\mathbf{C}}_{\mathbf{o}}\mathbf{V}\_\_\_\_\_\_\_\_\_\_\_\_\_\left(1\right)$

where$Q_o$ is the initial charge,$C_o$ is the initial capacitance and V is the potential difference between the two plates.

role="math" localid="1664255788557" $\mathbf{V}\mathbf{=}{\mathbf{E}}_{\mathbf{o}}\mathbf{d}\_\_\_\_\_\_\_\_\_\_\_\_\_\left(2\right)$

where V is the potential difference between the two plates,${\mathrm{E}}_{\mathrm{o}}$ is the electric field and d is the distance between the two plates

$\mathbf{C}\mathbf{=}{\mathbf{KC}}_{\mathbf{o}}\mathrm{\_}\_\_\_\_\_\_\_\_\_\_\_\_\left(3\right)$

where${\mathrm{C}}_{\mathrm{o}}$ is the initial capacitance, Cis the capacitance with the dielectric material and K is the dielectric constant

according to equation (1)

$\mathrm{Q}=\mathrm{CV}$

using equation (3)

$\mathbf{Q}\mathbf{=}{\mathbf{KC}}_{\mathbf{o}}\mathbf{V}\_\_\_\_\_\_\_\_\_\_\_\_\_\left(4\right)$

where, Q is the charge when a dielectric material is inserted between the two plates, K is the dielectric constant

(a)

We are given

the initial capacitance of a capacitor ${\mathrm{C}}_{\mathrm{o}}=8.00\mathrm{pF}=8.00\times {10}^{-12}\mathrm{F}$

the separated distance between the two plates is $\mathrm{d}=1.50\mathrm{mm}=0.0015\mathrm{m}$the electric field without a dielectric material.

using equation, (2)

the potential difference between the two plates

role="math" localid="1664255530284" $$\begin{gathered} \mathrm{V}={\mathrm{E}}_{\mathrm{o}}\mathrm{d} \\ =\left(3.00\times {10}^4\mathrm{V}/\mathrm{m}\right)\left(0.0015\mathrm{m}\right) \\ =45.0\mathrm{V} \end{gathered}$$

putting value of V in equation (1)

The charge on the two parallel plates

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{o}}={\mathrm{C}}_{\mathrm{o}}\mathrm{V} \\ =\left(8.00\times {10}^{-12}\mathrm{F}\right)\left(45.0\mathrm{V}\right) \\ =0.36\times {10}^{-9}\mathrm{C} \\ =0.36\mathrm{nC} \end{gathered}$$

Therefore, the maximum magnitude of the charge that can be placed on each plate is $0.36\mathrm{nC}$.

Using equation$\left(4\right)$

The charge Q when a dielectric material with a dielectric constant is inserted between the two plates

$$\begin{gathered} \mathrm{Q}={\mathrm{KC}}_0\mathrm{V} \\ =\left(2.70\right)\left(8.00\times {10}^{-12}\mathrm{F}\right)\left(45.0\mathrm{V}\right) \\ =0.97\times {10}^{-9}\mathrm{C} \\ =0.97\mathrm{nC} \end{gathered}$$

Therefore, the charge Qwhen a dielectric material with a dielectric constant is inserted between the two plates is 0.97nC .