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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.

User Guy Gordon
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1 Answer

1 vote

Answer:

Step-by-step 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{(3*2*10^(-3))/(1.9*10^(13)) }

t = 1.77 x 10⁻⁸ s

User HaneTV
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