Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charge density fo

Question

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charge density for each plate has magnitude 47.0 nC/m2, what is the magnitude of E⃗ in the region between the plates?

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Farah 4 years 2021-09-02T01:30:08+00:00 1 Answers 16 views 0

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  1. ThanhThu
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    2021-09-02T01:31:40+00:00
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    Answer:

    5.3\times 10^3 N/C

    Explanation:

    We are given that

    Distance between plates=d=2.2 cm=2.2\times 10^{-2} m

    1 cm=10^{-2} m

    \sigma=47nC/m^2=47\times 10^{-9}C/m^2

    Using 1 nC=10^{-9} C

    We have to find the magnitude of E in the region between the plates.

    We know that the electric field for parallel plates

    E=\frac{\sigma}{2\epsilon_0}

    E_1=\frac{\sigma}{2\epsilon_0}

    E_2=\frac{\sigma}{2\epsilon_0}

    E=E_1+E_2

    E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

    Where \epsilon_0=8.85\times 10^{-12}C^2/Nm^2

    Substitute the values

    E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}

    E=5.3\times 10^3 N/C

    Hence, the magnitude of E in the region between the plates=5.3\times 10^3 N/C

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