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A particle with a charge of +4.20 nC is in a uniform

Chapter 23, Problem 14E

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QUESTION:

A particle with a charge of +4.20 nC is in a uniform electric field 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 +1.50 × 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 end point? (c) What is the magnitude of .

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QUESTION:

A particle with a charge of +4.20 nC is in a uniform electric field 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 +1.50 × 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 end point? (c) What is the magnitude of .

ANSWER:

Solution 14E The work done in moving a charge q across a potential difference V is given by W = qV . From the data given, we can calculate the potential difference and then the electric field. Given, charge q = + 4.20 nC = + 4.20 × 10 9C Kinetic energy = 1.50 × 1

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