A charged particle enters a region with a uniform electric field that has equipo

Question

A charged particle enters a region with a uniform electric field that has equipotential lines as shown. The field slows the particle until it stops and reverses direction. When the particle passes through location A, which has an electric potential of 150 C, the particle has a kinetic energy of 645 eV. Location B, where the particle stops and turns around, has an electric potential of 120 V. What is the sign of the charge of the particle Positive Negative It cannot be determined from this information What is the magnitude of the charge of the particle 3.44e - 018 C 3.5e - 014 C 7.94e - 017 C 4.79e - 015 C 7.57e - 018 C

Explanation / Answer

14. The particle has initial velocity towards right, but slows down and stops at B and moves towards left. Imp-lies the particle experienced force to the left are repulsive force, we have –ve charge to the right and +ve charge to the left. Hence the particle is –vely charged and has a net force to the left.

15. when the particle reached 120V potential it stopped, the force at B is sufficient to stop the particle i.e. 120ev

Change in Potential from A to B = (150 -120) = 30 V

1 ev is the energy gained or lost by en electron when it passes through a potential difference of 1V

If an electron of charge (1.6e-19 C) passes through the potential difference of 30 V the KE will change by 30 ev.

The particle has lost an energy of 645 ev through a pot difference of 30V

Hence charge on the particle

Q = (645/30)*1.6e-19 C = 3.44e-18 C

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