A particle with a charge of 410 nC is in a uniform electric

A particle with a charge of +4.10 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left. After it has moved 6.00 cm, its kinetic energy is found to be +2.50 x 10\"6 J. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the endpoint? (c) What is the magnitude of f ?

Solution

a) The work done by electric force is exactly equal to the kinetic energy of the particle

W = 2.50 * 10^-6 J

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b) v_initial - v_final = W / q

= (2.50 * 10^-6) / (4.10 * 10^-9)

= 610 V

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c) E = V / d

= (610) / (0.06)

= 10166 N/C = 1.01 * 10⁴ N/C