Question

A particle beam is made up of many protons, each with a kinetic energy of 3.25times 10-15 J. A proton has a mass of 1.673 times 10-27 kg and a charge of 1.602 times 10-19 C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m?

Answer 1

Answer:

   E = 1.024 x 10⁴ V/m

Explanation:

given,

KE of the particle = 3.25 x 10⁻¹⁵ J

mass of the proton = 1.673 x 10⁻²⁷ Kg

charge of the proton = 1.602 x 10⁻¹⁹ C

distance = 2 m

Electric field = ?

we know

work done = KE, and also

W = q V

now,

$V = \dfrac{W}{q}$

$V = \dfrac{3.25\times 10^{-15}}{1.602\times 10^{-19}}$

  V = 2.028 x 10⁴ V

now, using equation of electric field

$E = \dfrac{V}{d}$

$E = \dfrac{2.028\times 10^4}{2}$

   E = 1.024 x 10⁴ V/m

hence, the magnitude of the electric field is equal to  E = 1.024 x 10⁴ V/m