Find the field produced by a uniformly charged sheet carrying \(87 \mathrm{pC} / \mathrm{m}^{2}\).

Short Answer

Expert verified
The electric field produced by the uniformly charged sheet is \(4.92 \mathrm{N/C}\).

Step by step solution

01

Convert charge to standard units

We need to convert the surface charge density from pC/m\(^2\) to C/m\(^2\). Given \(87 \mathrm{pC}/\mathrm{m}^{2}\), the conversion from pC (picocoulombs) to C (coulombs) is 1C = 1e12 pC. Hence, \(\sigma = 87 \times 10^{-12} \mathrm{C}/\mathrm{m}^{2}\).
02

Substitute values into formula

Substitute the values of \(\sigma\) and \(\epsilon_0\) into the formula \(E = \sigma / 2\epsilon_0\). The permittivity of free space, \(\epsilon_0\), is approximately \(8.85 \times 10^{-12} \mathrm{C}^{2}/\mathrm{N}\cdot\mathrm{m}^{2}\).
03

Calculate the electric field

Calculate the electric field by plugging in the values into the formula. The result is \(E = (87 \times 10^{-12}) / (2 \times 8.85 \times 10^{-12}) = 4.92 \mathrm{N/C}\). The field points away from the sheet if the charge is positive, and toward the sheet if the charge is negative.

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