Answer to Question #132397 in Mechanics | Relativity for Jessica

Question

A moving particle encounters an external electric field that decreases its kinetic energy from 9330 eV to 7410 eV as the particle moves from position A to position B. The electric potential at A is -45.0 V, and that at B is +17.0 V.

Determine the charge of the particle.

Include the algebraic sign (+ or -) with your answer.

Answer 1

from conservation of energy

$\Delta k=\Delta k_B-\Delta k_A=-q\Delta V=-q(V_b-V_A)$

where

K= kinetic energy(at and and at B)

q= charge

V= potential(at A and at B)

therefore making q subject of the formula

$q=-\frac{K_B-K_A}{V_b-V_a}=\frac{7410eV-9330eV}{17-(-45)}=-\frac{-1920eV}{62}=30.968eV$

$30.968eV=30.968e\times\frac{1.6\times 10^{-19}c}{e}$

$q=4.9549\times10^{-18}c$