Question

Parallel-plate capacitor has voltage 200 V, charge 0,2 micro

coulombs and distance between plates os 3 mm. Find the surface area
of the plates.
Give your answer in square meters to two decimal places.

Answer 1

The surface area of the plates of the parallel-plate capacitor is approximately 0.34 square meters when rounded to two decimal places.

Step-by-step explanation:

To find the surface area of the plates of a parallel-plate capacitor with a voltage of 200 V, a charge of 0.2 microcoulombs (or 0.2 x 10-6 C), and a distance between plates of 3 mm (or 0.003 m), we use the relationship between capacitance (C), charge (Q), and voltage (V), which is C = Q / V.

The capacitance of a parallel-plate capacitor is also given by C = ε0 A / d, where ε0 is the vacuum permittivity (ε0 = 8.85 x 10-12 F/m), A is area, and d is the separation between plates.

First, calculate the capacitance using the charge and voltage: C = Q / V = (0.2 x 10-6 C) / (200 V) = 1 x 10-9 F (1 nF). Then rearrange the capacitance formula to solve for A: A = C • d / ε0.

Substituting the known values: A = (1 x 10-9 F) • (0.003 m) / (8.85 x 10-12 F/m), which gives A ≈ 0.339 m2 or 0.34 m2 when rounded to two decimal places.