Question

A proton is released from rest in a uniform electric field. After the proton has traveled a distance of 10.0 cm, its speed is 1.4 x 10^6 m/s in the positive x direction.
Required:
a. Find the magnitude and direction of the electric field.
b. Find the speed of the proton

Answer 1

Answer:

Explanation:

distance travelled s = 10 cm

speed v = 1.4 x 10⁶ m /s

v² = u² + 2as

u = 0

v² = 2as

( 1.4 x 10⁶ )² = 2 x a x .10

a = 9.8 x 10¹² m /s²

force on proton = mass x acceleration

= 1.67 x 10⁻²⁷ x 9.8 x 10¹²

= 16.366 x 10⁻¹⁵ N .

If magnitude of electric field be E

force on proton

= E x charge on proton

= E x 1.6 x 10⁻¹⁹

E x 1.6 x 10⁻¹⁹ = 16.366 x 10⁻¹⁵

E = 10.22 x 10⁴ N/C

The direction of electric field will be positive x - direction .

b )

Speed of proton = 1.4 x 10⁶  m /s .