Question

Two large parallel conducting plates are 9.0 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 3.7 nC/m 2. Determine the magnitude of the electric potential difference between the plates.

Answer 1

The magnitude of the electric potential difference between the plates is 37.63 V.

Given:

Distance between the plates, d = 9 cm

Surface charge density, ρ = 3.7 nC/m² = 3.7 × 10⁻⁹ C/m²

Calculation:

For two parallel plate conductors, the electric field is given as:

E = ρ/ε₀

where ρ is the charge density

           ε₀ is the permeability of free space

Applying values in the above equation we get:

E = ( 3.7 × 10⁻⁹ C/m²)/(8.85 × 10⁻¹² C²/m)

  = 418.08 V/m

Now, the potential difference between the plates separated by a distance 'd' is given as:

ΔV = Ed

where E is the electric field

           d is the distance of separation between the plates

Applying values in the above equation we get:

ΔV = (418.08 V/m) × (0.09 m)

     = 37.63 V

Therefore, the magnitude of the electric potential difference between the plates is 37.63 V.

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