When a p.D of 150 qpplied to the plates of a parallel plate. The plate carry a surface charge density of 30nc/cm^2 what is the spacing bettwen the plates

d = 4.425 x 10⁻⁶ m = 4.425 μm

Explanation:

The charge on plates can be given as:

where,

d = spacing between plates = ?

ε₀ = Permitivity of free space  = 8.85 x 10⁻¹² C²/Nm²

σ = surface charge density = (30 nC/cm²)(10⁻⁹ C/1 nC)(1 cm²/10⁻⁴ m²)

σ = 3 x 10⁻⁴ C/m²

V = Potential Difference = 150 V

Therefore,

d = 4.425 x 10⁻⁶ m = 4.425 μm

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Explanation: