Question

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 286 N/C, find the following.
(a) Find the acceleration of the positron. m/s2
(b) Find the positron's speed after 8.70 × 10-9 s. Assume that the positron started from rest. m/s

Answer 1

Answer:

a) a = 5.03x10¹³ m/s²

b) $V_{f} = 4.4 \cdot 10^{5} m/s$

Explanation:    

a) The acceleration of the positron can be found as follows:

$F = q*E$    (1)

Also,

$F = ma$    (2)

By entering equation (1) into (2), we have:

$a = \frac{F}{m} = \frac{qE}{m}$

Where:

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C    

E: is the electric field = 286 N/C

$a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2}$

b) The positron's speed can be calculated using the following equation:

$V_{f} = V_{0} + at$

Where:

$V_{f}$: is the final speed =?

$V_{0}$: is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

$V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s$

I hope it helps you!