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A particle with charge +7.60 nC is in a uniform electric
Chapter 23, Problem 55P(choose chapter or problem)
A particle with charge is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved the additional force has done \(6.50 \times 10^{-5} \mathrm{~J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{~J}\) of kinetic energy.
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the end point?
(c) What is the magnitude of the electric field?
Questions & Answers
QUESTION:
A particle with charge is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved the additional force has done \(6.50 \times 10^{-5} \mathrm{~J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{~J}\) of kinetic energy.
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the end point?
(c) What is the magnitude of the electric field?
ANSWER:Introduction First we have to calculate the work done by the electric field. Then we have to calculate the potential difference and finally we have to calculate the magnitude of the electric field. Step 1 5 The total work done by the additional force is a = 6.50 × 10 J . The kinetic energy of the charge is E = 4.35 × 105 J. k Hence the work done by the electric field is W = 6.50 × 10 5 J 4.35 × 105 = 2.15 × 105J e