The cross sectional area of an optical fibre is 6.3 x 10^-6 m^2 The intensity of the light entering the optical fibre is 3.2 x 1
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Answer:
.
Explanation:
Since no external force is acting on the system.
Therefore, Total energy remains constant before and after.
So, Total energy of system= energy due to potential applied+kinetic energy
(Here v=velocity ,V=potential ,q=charge and m=mass).
Putting values .
We get, .
At point B charged particle is moving faster as compared to point A.
Hence, it is the required solution.