Answer:
The speed of the particle after travelling for 0.5 m is 100 m/s.
Explanation:
It is given that,
Mass of the particle, m = 0.01 kg
Net charge on the particle, q = -0.05 C
Electric field strength, E = 2000 V/C
Distance travelled by the particle, d = 0.5 m
The work done due to motion of the particle is balanced by the change in kinetic energy as :
[tex]Fd=\dfrac{1}{2}mv^2[/tex]
v is the speed of the particle
F is the electric force
[tex]qEd=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2qEd}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}[/tex]
v = 100 m/s
So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.
