Two large parallel conducting plates are 8.0 cm apart and carry equal but opposite charges on their facing surfaces.The magnitude of the surface charge density on either of the facing surfaces is 2.0 nC/m2.Determine the magnitude of the electric potential difference between the plates.

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Quizplus · Accepted Answer

The electric field (E) between two parallel plates is given by E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space (8.85 x 10^-12 C²/N·m²). The potential difference (V) is then V = E * d, where d is the separation between the plates. Substituting the given values, V = (2.0 x 10^-9 C/m²) / (8.85 x 10^-12 C²/N·m²) * 0.08 m = 18 V.

Two Large Parallel Conducting Plates Are 8