Question
Two parallel plates of area $100 \mathrm{~cm}^{2}$ are given charges of equal magnitudes $8.9 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the dielectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{~V} / \mathrm{m} .$ (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
Step 1
Step 1: We know that the electric field E is given by the formula $E = \frac{Q}{\varepsilon_0 A}$, where Q is the charge, $\varepsilon_0$ is the permittivity of free space, and A is the area. Show more…
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Two parallel plates of area $100 \mathrm{~cm}^{2}$ are given charges of equal magnitudes $8.4 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the dielectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{~V} / \mathrm{m}$. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
Two parallel plates of area 100 $\mathrm{cm}^{2}$ are given charges of equal magnitudes $8.9 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the diclectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{V} / \mathrm{m} .$ (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
Two parallel plates of area $100 \mathrm{~cm}^{2}$ are given excess charges of equal amounts $8.9 \times 10^{-7} \mathrm{C}$ but opposite signs. The electric field within the dielectric material filling the space between the plates is $1.4 \times 10^{6} \mathrm{~V} / \mathrm{m} .$ (a) Calculate the dielectric constant of the material. (b) Determine the amount of bound charge induced on each dielectric surface.
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