Question

An electron is released with a velocity of 5×10^6m/sec in an electric field of 10^3N/C which has been applied so as to opposite it's motion. What distance would electron travel . And also calculate how much time could it take before it broad to rest

Answer 1

Given info : electron is released with a velocity of 5 × 10⁶ m/s in an electric field of 10³ N/C which has been applied so as to opposite its motion.

To find : What distance would electron travel. And also time taken to become rest.

Solution : velocity of electron, v = 5 × 10⁶ m/s

Electric field intensity , E = 10³ N/C

Using work energy theorem,

Work done by electric field on electron to stop it = kinetic energy of electron

⇒eEx = 1/2 mv²

Here m is mass of electron i.e., m = 9.1 × 10¯³¹ Kg and e is charge of electron i.e., e = 1.6 × 10^-19 C

Now, 1.6 × 10^-19 × 10³ x = 1/2 × 9.1 × 10¯³¹ × (5 × 10⁶)²

⇒1.6 × 10¯¹⁶ × x = 4.55 × 25 × 10^-19

⇒x = (4.55 × 25/1.6) × 10¯³

= 71.094 × 10¯³ m = 71.094 mm

distance travelled by electron is 71. 094 mm

Acceleration, a = eE/m

= (1.6 × 10^-19 × 10^3)/(9.1 × 10^-31)

= (1.6/9.1) × 10¹⁵

= 1.758 × 10¹⁴ m/s²

So time taken is given by, s = 1/2 at²

⇒t = √{2 × 71.094 × 10^-3/1.758 × 10¹⁴}

= 2.84 × 10^-8 s

= 28.4 ns

Therefore the time taken is 28.4 ns