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

Answer:

d = 4.425 x 10⁻⁶ m = 4.425 μm

Explanation:

The charge on plates can be given as:

$q = \sigma A=CV\\where,\\\\C = Capacitance\ of\ parallel\ plate\ capacitor = \frac{\epsilon_oA}{d}\\\\therefore,\\\\\sigma A=(\frac{\epsilon_oA}{d})V\\\\d = (\frac{\epsilon_o}{\sigma})V$

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,

$d = \frac{(8.85\ x\ 10^{-12}\ C^2/Nm^2)}{3\ x\ 10^{-4}\ C/m^2}(150\ V)$

d = 4.425 x 10⁻⁶ m = 4.425 μm