Asked by Mike
A moving particle encounters an external electric field that decreases its kinetic energy from 9530 eV to 7800 eV as the particle moves from position A to position B. The electric potential at A is -69.0 V, and that at B is +39.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.
Answers
Answered by
bobpursley
ke change=9530-7800 figure that out in eV
charge*69=-kechange above.
divide both sides by 69, you get a charge in e.
multiply by e to get the charge in couloumbs.
charge*69=-kechange above.
divide both sides by 69, you get a charge in e.
multiply by e to get the charge in couloumbs.
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