Question

Two parallel conducting plates are oriented in an ? − ? coordinate system. The plates are separated by a distance ? = 0.05 ?, and the origin of the coordinate system is halfway between the plates. The plate separation is very small compared to the size of the plates. The left plate is at a potential ?_L = 50 ?, and the right plate is at a potential ?_R = 10 ?.

1. A proton at the origin begins at rest. Use conservation of energy to find its velocity when it hits one of the plates. Which plate does it hit?

2. An electron at the origin begins with an initial velocity ?⃗? = (4 × 106 ?/?)?̂. Use conservation of energy to find its velocity when it hits the right plate?

3. What is the acceleration of a proton at the origin? Express your answer in unit vector notation.

3. What is the acceleration of a proton at the origin? Express your answer in unit vector notation.

Answer 1

1.

V_L = Potential on the left plate = 50 Volts

V_R = Potential on the right plate = 10 Volts

d = distance between the plates = 0.05 m

E = magnitude of electric field between the plates

Electric field is given as

E = (V_L - V_R )/d = (50 - 10)/0.05 = 800 V/m

Electric field is direction from high to low potential, hence from left to right

Proton moves in the direction of electric field.

x = distance moved by proton from origin to right plate = d/2 = 0.05/2 = 0.025 m

v_o = initial speed of proton = 0 m/s

v = final speed of proton as it hits the right plate = ?

m_p = mass of proton = 1.67 x 10^-27 kg

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

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy lost

(0.5) m_p (v² - v_o²) = q (E x)

(0.5) (1.67 x 10^-27) (v² - (0)²) = (1.6 x 10^-19) (800 x 0.025)

v = 6.2 x 10⁴ m/s

= 6.2 x 10⁴ m/s

b)

V_L = Potential on the left plate = 50 Volts

V_R = Potential on the right plate = 10 Volts

d = distance between the plates = 0.05 m

E = magnitude of electric field between the plates

Electric field is given as

E = (V_L - V_R )/d = (50 - 10)/0.05 = 800 V/m

Electric field is direction from high to low potential, hence from left to right

Electron moves in the opposite direction of electric field.

x = distance moved by electron from origin to right plate = d/2 = 0.05/2 = 0.025 m

v_o = initial speed of electron = 4 x 10⁶ m/s

v = final speed of electron as it hits the right plate = ?

m_e = mass of electron = 9.1 x 10^-31 kg

q = charge on electron = 1.6 x 10^-19 C

Using conservation of energy

Kinetic energy lost by electron  = Electric potential energy gained

(0.5) m_e (v_o² - v²) = q (E x)

(0.5) (9.1 x 10^-31) ((4 x 10⁶)² - v²) = (1.6 x 10^-19) (800 x 0.025)

v = 3 x 10⁶ m/s

= 3 x 10⁶ m/s

c)

= acceleration of proton at origin = (qE/m_p ) = (1.6 x 10^-19) (800 ) /(1.67 x 10^-27) = 7.7 x 10¹⁰ m/s²

d)

= acceleration of electron at origin = - (qE/m_e ) = (1.6 x 10^-19) (800 ) /(9.1 x 10^-31) = - 1.4 x 10¹⁴ m/s²