Question

rt A How much work does the electric motor of a Van de Graaff generator do to lift a positive ion (q=e) if the potential of the spherical electrode is 2.0 MV ?

Answer 1

Answer:

W = 2.0 MeV = 3.2*10⁻¹³ J

Explanation:

• Assuming that the ion is located originally at a point where the electrostatic potential is zero, the change in energy of the ion is just the change in electrostatic potential times the charge of the ion, as follows:

        $\Delta E = e*V = 1.6e-19 C * 2.0e6 V = 2.0 MeV = 3.2e-13 J$

• This change in energy, assuming no friction present, must be equal to the work done by the motor on the ion:

        W = 2.0 MeV = 3.2*10⁻¹³ J.