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 326 N/C, find the following.(a) Find the acceleration of the positron. m/s2 (b) Find the positron's speed after 8.70 Ã 10-9 s. Assume that the positron started from rest. m/s

Answer 1

Answer:

(a) 5.73 * 10¹³ m/s²

(b) 4.99 * 10⁵ m/s

Explanation:

(a) We know that Electrical force is the product of electric charge and electric field:

F = qE

We also know that Force is the product of mass and acceleration:

F = ma

Therefore, equating both of them:

ma = qE

Acceleration, a, will be:

a = (qE) / m

Electric charge of a positron = 1.602 * 10^(-19) C

Mass of a positron = 9.11 * 10^(-31) kg

Acceleration will be:

a = (1.602 * 10^(-19)) * 9.11 * 10^(-31)) / 326

a = 5.73 * 10¹³ m/s²

(b) Acceleration is the time rate of change of velocity. It is given as:

a = v/t

Velocity, v, will then be:

v = a * t

Time, t = 8.7 * 10^(-9) s

Therefore, velocity will be:

v = 5.73 * 10^(13) * 8.7 * 10^(-9)

v = 49.85 * 10⁴ m/s = 4.99 * 10⁵ m/s