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3 years ago

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To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

th the same mass as an electron but the opposite charge (+e). If a positron is accelerated by a constant electric field of magnitude 343 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

1 answer:

Gnom [1K]3 years ago

6 0

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

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

Substitute the values of q, E and m in the above equation.

$a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }$

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

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