A movig particle encounters an external electric field that decreases its kinetic energy from 9670 eV to 7540 eV as the particle moves from position A to position B. The electric potential at A is -66.0 V, and that at B is +16.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.
physics - Elena, Saturday, January 19, 2013 at 6:42am
ΔKE = 9670 – 7540 =2130 eV =2130•1.6•10⁻¹⁹= 3.41•10⁻¹⁶ J
Δφ=|φ(A)-φ(B)| = 16 –(-66) = 82 V
ΔKE =q• Δφ
q =ΔKE/ Δφ =3.41•10⁻¹⁶/82=4.2•10⁻¹⁸ C
The energy of the charge is decreasing at the motion from the negative to the positive potential => the sign is negative