Question

A set of charged plates is

Answer 1

Answer:

2.39×10¯⁵ m²

Explanation:

From the question given above, the following data were obtained:

Distance (d) = 8.08×10¯⁵ m

Charge (q) = 2.24×10¯⁹ C

Potential difference = 855 V

Area(A) =?

NOTE: Permittivity (ε₀) = 8.854×10¯¹² F/m

The area can be obtained as follow:

q = ε₀AV/d

2.24×10¯⁹ = 8.854×10¯¹² × A × 855 / 8.08×10¯⁵

2.24×10¯⁹ = 7.57×10¯⁹ × A / 8.08×10¯⁵

Cross multiply

7.75×10¯⁹ × A = 2.24×10¯⁹ × 8.08×10¯⁵

7.75×10¯⁹ × A = 1.81×10¯¹³

Divide both side by 7.75×10¯⁹

A = 1.81×10¯¹³ / 7.75×10¯⁹

A = 2.39×10¯⁵ m²

Thus, the area of the plate is 2.34×10¯⁵ m²