Questions: Two large parallel metal plates are uniformly and oppositely charged and the electric field between them is 7.6 × 10^6 N / C. What is the charge density on each plate? All answers in μC / m^2

Two large parallel metal plates are uniformly and oppositely charged and the electric field between them is 7.6 × 10^6 N / C. What is the charge density on each plate? All answers in μC / m^2
Transcript text: Two large parallel metal plates are uniformly and oppositely charged and the electric field between them is $7.6 \times 10^{6} \mathrm{~N} / \mathrm{C}$. What is the charge density on each plate? All answers in $\mu \mathrm{C} / \mathrm{m}^{2}$
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Solution

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Solution Steps

Step 1: Understand the Relationship Between Electric Field and Charge Density

The electric field \( E \) between two large parallel plates is related to the surface charge density \( \sigma \) on the plates by the equation: \[ E = \frac{\sigma}{\varepsilon_0} \] where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \).

Step 2: Solve for Charge Density

Rearrange the equation to solve for the surface charge density \( \sigma \): \[ \sigma = E \cdot \varepsilon_0 \] Substitute the given electric field \( E = 7.6 \times 10^{6} \, \mathrm{N/C} \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \) into the equation: \[ \sigma = 7.6 \times 10^{6} \, \mathrm{N/C} \times 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2 \]

Step 3: Calculate and Convert to \(\mu \mathrm{C} / \mathrm{m}^{2}\)

Calculate the value of \( \sigma \): \[ \sigma = 67.26 \times 10^{-6} \, \mathrm{C/m}^2 \] Convert this to \(\mu \mathrm{C} / \mathrm{m}^{2}\): \[ \sigma = 67.26 \, \mu \mathrm{C/m}^2 \] Since the options are rounded, the closest answer is \(67 \, \mu \mathrm{C/m}^2\).

Final Answer

\(\boxed{67}\)

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