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Grade 12Electrostatics

A charged particle of charge 2×10^-6C and mass 10 milligram ,moving with a velocity of 1000m/s enters a uniform electric field of strength 10³ NC^-1directed perpendicular to its direction of motion . Find the velocity and displacement of the particle after 10 s.

Swathi

8 Years agoGrade 12

1 Answer

Eshan

8 Years ago

Dear student,

The component of velocity perpendicualr to the direction of electric field will remain constant since the particle does not accelerate in that direction.
Hence $v_{perpendicular}=1000m/s \implies s_{perpendicular}=v_{perpendicular}t=10000m$

The acceleration of particle in direction of electric field= $\dfrac{qE}{m}=\dfrac{2\times 10^{-6}\times 10^3}{0.01}m/s^2=0.2m/s^2$

Hence the velocity of particle parallel to electric field, $v_{parallel}=at=2m/s$
$\implies s_{parallel}=\dfrac{1}{2}at^2=10m$

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