Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.40×105 V/m .

Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.40×105 V/m . When the space is filled with dielectric, the electric field is E= 2.20×105 V/m . Part A What is the charge density on each surface of the dielectric?

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Thông Đạt 5 years 2021-08-22T10:26:31+00:00 1 Answers 26 views 0

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  1. Jezebel
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    2021-08-22T10:28:23+00:00
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    Answer:

    \sigma_i=1.06*10^{-6}C

    Explanation:

    When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:

    \sigma_i=\sigma(1-\frac{E}{E_0})

    Where E is the eletric field with dielectric and E_0 is the electric filed without it. Recall that \sigma is given by:

    \sigma=\epsilon_0E_0

    Replacing this and solving:

    \sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C

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