Question

lets work on problem 1 from lecture notes 4 as our example. A particle with a mass of 5.00 grams and a charge of 4 microcoulombs has a speed of 0.75 m/s when it passes through a point at which the potential is -1200 volts electron and a proton are released form rest in a uniform electric field that has a magnitude of 500 N/C. The energy and speed of each particle is measured after it has moved through a distance of 25 cm. Assume the particles do not influence one another, but are influenced only by the electric field. Without doing any calculations, determine which particle (a) has more kinetic energy, (b) has a higher speed, (c) takes more time to cover 25 cm. Justify your answers. Now calculate the kinetic Energy, speed and elapsed time for (d) the electron and (e) the proton.

Answer 1

The force acting on each particle is the same because the electric field is uniform everywhere and the charge on each particle is the same. However, the direction is *opposite* because positive and negative charges are either attracted to the source of the electric field or repulsed by it.

Electron, Fe = EQ = (500 N/C)(-1.60x10^-19 C) = - 8 x10^-17 N

Proton, Fp = EQ = (500 N/C)(1.60x10^-19 C) = 8 x 10^-17 N

Note that even though the force is the same on each particle, the acceleration will NOT be the same. This is because the proton is much more massive than an electron (it is easier for the force to speed up the electron).

By Newton's law: For electron, Fe = m_ea_e;

(- 8 x10^-17 N)= (9.1x10^-31 kg)a_e

a_e = - 8.79 x10^13 m/s^2

For Proton, Fp = m_pa_p;

(8 x10^-17 N)= (1.67x10^-27 kg)a_p

a_p = 4.79x10^10 m/s^2

we know that v²-u²=2as

For Electron,

v_e² = (0 m/s) + 2*( - 8.79 x10^13 m/s^2)*(25x10^-2 m)

v_e= - 6.63 x 10⁶ m/s

(the negative sign simply means that the velocity is in the opposite direction for the electron when compared to the proton)

For Proton, v_p²  = 0 +2*(4.79x10^10 m/s^2 )*(25x10^-2 m)

v_p = 1.55x 10⁵ m/s

K.E. of electron = (1/2)m_ev_e² = (1/2)*9.1x10^-31 *(6.63 x 10⁶)^2 = 18.85 eV

K.E. of Proton = (1/2)m_pv_p² = (1/2)*1.67x10^-27*(1.55 x 10⁵)^2 = 0.08 eV

So (a) electron has more kinetic energy,

(b) electron has a higher speed

(c) proton will take more time to cover 25 cm

v=u+at

t = (v-u)/a = v/a, bcoz, u=0

for electron time elapsed, t_e = v_e/a_e

t_e = 6.63 x 10⁶/8.79 x10¹³ = 75.4 ns

for proton time elapsed, t_p = v_p/a_p

t_p =1.55x 10⁵ /4.79x10¹⁰ = 3236 ns