Question

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each plate has a charge density of magnitude 5.0 pC/m^2?
1.02 millivolts 1.43 millivolts 2.26 millivolts 3.34 millivolts 4.43 millivolts

Answer 1

The potential difference across the parallel plate capacitor is 2.26 millivolts

Capacitance of a parallel plate capacitor

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

• ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
• A = area of plates and
• d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

Charge on plates

Also, the surface charge on the capacitor Q = σA where

• σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
• a = area of plates.

The potential difference across the parallel plate capacitor

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

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