Two parallel plates having charges of equal magnitude but opposite sign are sepa

Question

Two parallel plates having charges of equal magnitude but opposite sign are separated by 25.0 cm. Each plate has a surface charge density of 40.0 nC/m2. A proton is released from rest at the positive plate.

(a) Determine the magnitude of the electric field between the plates from the charge density.
kN/C

(b) Determine the potential difference between the plates.
V

(c) Determine the kinetic energy of the proton when it reaches the negative plate.
J

(d) Determine the speed of the proton just before it strikes the negative plate.
km/s

(e) Determine the acceleration of the proton.
m/s2 towards the negative plate

(f) Determine the force on the proton.
N towards the negative plate

(g) From the force, find the magnitude of the electric field.
kN/C

(h) How does your value of the electric field compare with that found in part (a)?

Explanation / Answer

Here ,

seperation , d = 25 cm = 0.25 m

surface charge density , sigma = 40 nC/m^2

a)

magnitude of electric field = sigma/epsilon_0

magnitude of electric field = 40 *10^-9/(8.854 *10^-12)

magnitude of electric field = 1129 N/C

the magnitude of electric field is 1129 N/C

b)

potential difference = electric field * distance

potential difference = 1129 * 0.25 V

potential difference = 282.4 V

c) for the kinetic energy of proton

kinetic energy of proton = V * charge

kinetic energy of proton = 282.4 * 1.602 *10^-19 J

kinetic energy of proton = 4.523 *10^-17 J

d)let the speed of proton is v

0.5 * m * v^2 = kinetic energy of proton

0.5 * 1.67 *10^-27 * v^2 = 4.523 *10^-17

solving for v

v = 9.97 * 10^6 m/s

the speed of electron is 9.97 * 10^6 m/s

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