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Numerical · Q26

Q.An infinite plane sheet carries a uniform surface charge density σ=8.85×10−8 C/m2\sigma = 8.85\times10^{-8}\ \text{C/m}^2. Using Gauss's theorem, find the magnitude of the electric field at any point near the sheet (take ϵ0=8.85×10−12 C2N−1m−2\epsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}).

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Given σ=8.85×10−8 C/m2\sigma=8.85\times10^{-8}\ \text{C/m}^2, ϵ0=8.85×10−12 C2N−1m−2\epsilon_0=8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}. Using the Gauss's-theorem result of Section 1.16:

E=σ2ϵ0=8.85×10−82×8.85×10−12=8.85×10−81.77×10−11=5×103 N/CE = \frac{\sigma}{2\epsilon_0} = \frac{8.85\times10^{-8}}{2\times8.85\times10^{-12}} = \frac{8.85\times10^{-8}}{1.77\times10^{-11}} = 5\times10^3\ \text{N/C} …

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