Question

A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis? (b) How long will it take for the particle to travel a distance of 40 cm? (c) What will be the speed of the particle after travelling this distance? (d) How much is the work done by the electric force on the particle during this period?

Answer 1

(a) Electric force and the force of gravity is $F=3 N$ and $q=9.8 \times 10^{-3}$ .

(b) Time taken for the particel to travel a distance of 40 cm is, $t=0.01632$.

(c) Speed of the particle after travelling this distance is, $v=48.98 \mathrm{m} / \mathrm{s}$.

(d) Work done by the electric force is, w = 1.2 J .

Explanation:

A particle of mass, m = 1 g

Converting mass gram to kilogram  :

           $\text { Mass }=\frac{1}{1000}=10^{-3} \mathrm{kg}=0.001 \mathrm{kg}$

Initial velocity u = 0  

Charge q  $=2.5 \times 10^{-4} \mathrm{C}$

Electric field  E $=1.2 \times 10^{4} \mathrm{N} / \mathrm{c}$

S = 40 cm  

Cm to meter :

             $S=4 \times 10^{-1} \mathrm{m}=0.4 \mathrm{m}$

(a)  Electric force and the force of gravity :

            F = q E

           $F=2.5 \times 10^{-4} \times 1.2 \times 10^{4}$

           $F=3 N$

     So,  

           Force = mass × acceleration

                   F = ma

                   $a=\frac{E}{a}$

                   $a=\frac{3}{10^{-3}}$

                   $a=3 \times 10^{3}$

                   q = mg  

                   $q=9.8 \times 10^{-3}$          

(b) Time taken to travel a distance of 40 cm :

                    $S=u+\frac{1}{2} a t^{2}$

                    $S=0+\frac{1}{2} a t^{2}$

                    $S=\frac{1}{2} a t^{2}$

                   $2 S=a t^{2}$

                     $t^{2}=\frac{2 S}{a}$

                      $t=\sqrt{\frac{2 S}{a}}$

                      $t=\sqrt{\frac{2 \times 0.4}{3 \times 10^{3}}}$

                       $t=\sqrt{\frac{0.8}{3 \times 10^{3}}}$

                       $t=\sqrt{0.266 \times 10^{-3}}$  

                       $t=0.01632$

(c) Speed of the particle after travelling this distance :

                       $v^{2}=u^{2}+2 a s$

                       $v^{2}=0+2 \times 3 \times 10^{3} \times 0.4$

                       $v^{2}=6 \times 10^{3} \times 0.4$

                       $v^{2}=2.4 \times 10^{3}$

                         $v=\sqrt{2400}$

                         $v=48.98 \mathrm{m} / \mathrm{s}$

(d) Electric force work done :

                 w = F→td  

                 w = 3 × 0.4  

                  w = 1.2 J