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
B
6×106J
C
6×106J
D
6×106erg

Solution

The correct option is D $$6\times 10^{-6}$$JSince 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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