Question

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

Answer 1

d = 4.425 x 10⁻⁶ m = 4.425 μm

Step-by-step explanation:

The charge on plates can be given as:

`q = \sigma A=CV\\where,\\\\C = Capacitance\ of\ parallel\ plate\ capacitor = (\epsilon_oA)/(d)\\\\therefore,\\\\\sigma A=((\epsilon_oA)/(d))V\\\\d = ((\epsilon_o)/(\sigma))V`

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 = ((8.85\ x\ 10^(-12)\ C^2/Nm^2))/(3\ x\ 10^(-4)\ C/m^2)(150\ V)\\`

d = 4.425 x 10⁻⁶ m = 4.425 μm