Question

A particle of unknown charge obtains 0.06 J of kinetic energy when it moves from point A to point B. Point A has an electric potential of 888.0 V and point B has an electric potential of 333.0 V. Determine the magnitude and the sign of charge. [3]
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Answer #1

From law of conservation of energy

change in kinetic energy = change in potential energy

0.06 J =q*delta V

0.06 J =q*(vB - vA)

0.06 J =q*(333.0 V - 888.0 V )

q =(0.06 J)/(333.0 V - 888.0 V)

= -1.08108108*10^-4 C

= -108.108108 micro Coulomb

= -108 micro coulomb

magnitude q = 108 micro coulomb

sign : negative

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