Two parallel conducting plates are separated by 1.0 mm and carry equal but opposite surface charge densities. If the potential difference between them is 2.0 V, what is the magnitude of the surface charge density on each plate? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²)

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Quizplus · Accepted Answer

The potential difference between the plates is given by:$$V = Ed$$where:* V is the potential difference* E is the electric field* d is the distance between the platesThe electric field between the plates is given by:$$E = \frac{\sigma}{\varepsilon_0}$$where:* σ is the surface charge density* ε₀ is the permittivity of free spaceSubstituting the expressions for E and d into the first equation, we get:$$V = \frac{\sigma d}{\varepsilon_0}$$Solving for σ, we get:$$\sigma = \frac{\varepsilon_0 V}{d}$$Plugging in the given values, we get:$$\sigma = \frac{(8.85 imes 10^{-12} ext{ C}^2/ ext{N} \cdot ext{m}^2)(2.0 ext{ V})}{1.0 imes 10^{-3} ext{ m}} = 18 ext{ nC/m}^2$$

Two Parallel Conducting Plates Are Separated by 1