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Question

A particle having mass $$1$$g and electric charge $$10^{-8}$$C travels from a point A having electric potential $$600$$V to the point B having zero potential. What would be the change in its kinetic energy?


A
6×106erg
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B
6×106J
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C
6×106J
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D
6×106erg
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Solution

The correct option is D $$6\times 10^{-6}$$J
Since the particle will move under conservative force (Electric Force), it's mechanical energy will be consrved,So, change in mechanical energy will be equal to change in it's potential energy.
$$\Delta K.E$$ = $$\Delta P.E$$ = $$Q\Delta V$$ = $$10^{-8}\times 600 J$$ = $$ 6\times 10^{-6} J$$.And the change in kinetic energy will be positive because potential energy is decreasing so kinetic energy will be increased in order to conserve the mechanical energy.
Therefore, C is the correct option.

Physics

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