Question

A parallel-plate capacitor consists of two plates, each with an area of 27 cm^2 separated by 3.0 mm. The charge on the capacitor is 4.8 nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate? Steps please with right answer.

Answer 1

Answer:

Explanation:

Capacity of a parallel plate capacitor  C = ε₀ A/ d

ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.

C =(  8.85 X10⁻¹² X  27 X 10⁻⁴ ) / 3 X 10⁻³

= 79.65 X 10⁻¹³ F.

potential diff between plate = Charge / capacity

= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³

= 601 V

Electric field = V/d

= 601 / 3 x 10⁻³

= 2 x 10⁵ N/C

Force on proton

= charge x electric field

1.6 x 10⁻¹⁹ x 2 x 10⁵

= 3.2 x 10⁻¹⁴

Acceleration a = force / mass

= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷

= 1.9 x 10¹³ m s⁻²

Distance travelled by proton = 3 x 10⁻³

3 x 10⁻³ = 1/2 a t²

t = $\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }$

t = 1.77 x 10⁻⁸ s