Question

An electron moving to the right at v = 5.5×105 m/s enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of x = 5.0 cm. (a) What direction is required for the electric field? (b) What is the strength of the field?

Answer 1

Answer

a) Electron is moving toward right and brought to rest it means that force should be in opposite direction of motion electron.

We know F  = qE  here q = -e  so, Force will be F = - eE.

Force on electron will be acting in opposite direction so, the electric field direction will be towards Right.

b) we know,

   Force on electron = Electric force

         m a = eE

      $a = \dfrac{eE}{m}$

using equation of motion

 v² = u² + 2 a s

 0 = (5.5 x 10⁵)² - 2 x a x 0.05

$0.1\times \dfrac{eE}{m}= (5.5 \times 10^5)^2$

e = 1.6 x 10⁻¹⁹ C

mass of electron, m = 9.1.09 x 10⁻³¹ Kg

now,

$0.1\times \dfrac{1.6\times 10^{-19}\times E}{9.109\times 10^{-31}}= (5.5 \times 10^5)^2$

E = 17.22 N/C

hence, strength of electric field is equal to E = 17.22 N/C