Question

A proton with an initial velocity of 3.9 106 m/s enters a parallel-plate capacitor at an angle of 46° through a small hole in the bottom plate as shown in the figure below. If the plates are separated by a distance
d = 2.1 cm,
what is the magnitude of the minimum electric field to prevent the proton from hitting the top plate?
___ N/C

Compare your answer with the results obtained if the proton was replaced with an electron and choose an answer:

A. The magnitude of the electric field needed for a proton is larger than what is needed for an electron.

B. The magnitude of the electric field needed for a proton is smaller than what is needed for an electron.   

C. The magnitude of the electric field needed for a proton is the same as what is needed for an electron.

Answer 1

Given that :

initial velocity of proton, v₀ = 3.9 x 10⁶ m/s

angle, = 46 degree

separation distance between plate, d = 2.1 cm = 0.021 m

initial vertical velocity which given as, v_0,y = v sin                                                                { eq.1 }

inserting the values in eq.1,

v_0,y = (3.9 x 10⁶ m/s) sin (46⁰)

v_0,y = 2.8 x 10⁶ m/s

magnitude of the minimum electric field to prevent the proton from hitting the top plate which will be given as ::

using an equation, F = q E

or E = F / q                                                                                        { eq.2 }

from newton's second law of motion, F = m a

then E = m a / q                                                                                     { eq.3 }

where, m = mass of proton = 1.67 x 10^-27 kg

q = charge on proton = +1.6 x 10^-19 C

now, using equation of motion 3 to find 'a' :

v_y² = v_0,y² + 2 a d                                                                                   { eq.4 }

where, v_y = final vertical velocity = 0 m/s

inserting the values in eq.4,

(0 m/s)² = (2.8 x 10⁶ m/s)² + 2 a (0.021 m)

(7.84 x 10¹² m²/s²) = (0.042 m) a

a = 186.6 x 10¹² m/s²

a = 1.86 x 10¹⁴ m/s²

inserting these values in eq.4,

E = (1.67 x 10^-27 kg) (1.86 x 10¹⁴ m/s²) / (+1.6 x 10^-19 C)

E = (3.1 x 10^-13 N) / (+1.6 x 10^-19 C)

E = 1.93 x 10⁶ N/C

Compare your answer with the results obtained if the proton was replaced with an electron and choose an answer :

A. The magnitude of the electric field needed for a proton is larger than what is needed for an electron.

for an electron, E = 10.5 x 10² N/C