An electron (charge e = - 1.6 Times 10^-19 C, mass m_c = 9 Times 10^-31 kg) ente

Question

An electron (charge e = - 1.6 Times 10^-19 C, mass m_c = 9 Times 10^-31 kg) enters a uniform electric field (figure 2). The vector of the initial velocity is directed along the electric field lines. The electron is moving through a potential differences Delta V = -100 V before it stops. What is the direction of the electric force on the electron in Figure 2? What was the change of kinetic energy and the initial velocity v_o of the electron? Use the law of conservation of energy Delta K = -Delta U and a general expression for the kinetic energy K = mv^2/2

Explanation / Answer

a) towards left (-x axis)

b) change of kinetic energy = delta U

= q*delta_V

= -e*100 V

= -100 eV

= -100*1.6*10^-19 J

= -1.6*10^-17 J

c) K1 = 0.5*m*vo^2

==> vo = sqrt(2*KE/m)

= sqrt(2*1.6*10^-17/(9.1*10^-31))

= 5.93*10^6 m/s

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