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 .

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 × 10 6 J (a) Here, the kinetic energy of the particle is manifested as the work done on it. The particle gets its kinetic energy because of the work done on it by the electric force. Hence 6 the magnitude of the work done is 1.50 × 10 J . (b) The potential difference can be calculate from the equation W = qV . Therefore, the 6 9 potential difference V = W/q = 1.50 × 10 /4.20 × 10 V = 357 V (c) Distance moved by the particle d = 6.00 cm = 6.00 × 10 2 m Therefore, the magnitude of electric field = 357 2V /m = 5950 V /m 6.00×10