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. $-6\times {10}^{-6}$erg
• B. $-6\times {10}^{-6}$J
• C. $6\times {10}^{-6}$J
• D. $6\times {10}^{-6}$erg

Answer 1

Answer: (C)

$6\times {10}^{-6}$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.

$\Delta K.E$ = $\Delta P.E$ = $Q\Delta V$ = ${10}^{-8}\times 600J$ = $6\times {10}^{-6}J$.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.