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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 -

    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

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