Question

A parallel-plate capacitor has a Japacitance of 1.97 pF and a plate area of 5.86 cm². What is the separation distance between the plates?

a.2.63 x 10⁻¹¹ m
b.2.63 mm
c.15.4 mm
d.0.263 m
e.1.54 m

Answer 1

The capacitance (C) of a parallel-plate capacitor is given by the formula: C =
`(\varepsilon_0 \cdot A)/(d)` . The separation distance between the plates is 2.63 × 10⁻³ m or 2.63 mm. The correct answer is option (b) 2.63 mm.

Step-by-step explanation:

The capacitance (C) of a parallel-plate capacitor is given by the formula:

C =
`(\varepsilon_0 \cdot A)/(d)`

Where:

- C is the capacitance,

- ε₀ is the permittivity of free space (8.85 × 10⁻¹² F/m,

- A is the plate area,

- d is the separation distance between the plates.

Rearrange the formula to solve for d:

d =
`(\varepsilon_0 \cdot A)/(C)`

Plug in the given values:

d =
`\frac{(8.85 * 10^(-12) \, \text{F/m}) \cdot (5.86 * 10^(-4) \, \text{m}^2)}{1.97 * 10^(-12) \, \text{F}}`

d ≈ 2.63 × 10⁻³ m or 2.63 mm.

So, the correct answer is option (b) 2.63 mm.