A set of charged plates is separated by 2.22*10^-4 m. When
5.24*10^-9 C of charge is placed
on the plates, it creates a potential
difference of 240 V. What is the
area of the plates?
(The answer is *10^-4 m^2. Just fill
in the number, not the power.)

5.49×10¯⁴ m²

Explanation:

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

Distance (d) = 2.22×10¯⁴ m

Charge (Q) = 5.24×10¯⁹ C

Potential difference (V) = 240 V

Permittivity of free space (ε₀) = 8.85×10¯¹² F/m

Area (A) =?

Thus, the area of the plate can be obtained as follow:

Q = ε₀AV /d

5.24×10¯⁹ = 8.85×10¯¹² × A × 240 / 2.22×10¯⁴

5.24×10¯⁹ = 2.12×10¯⁹ × A / 2.22×10¯⁴

Cross multiply

2.12×10¯⁹ × A = 5.24×10¯⁹ × 2.22×10¯⁴

Divide both side by 2.12×10¯⁹

A = (5.24×10¯⁹ × 2.22×10¯⁴) / 2.12×10¯⁹

A = 5.49×10¯⁴ m²

Thus, the area of the plates is 5.49×10¯⁴ m².

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down