1) A charge Q 1 = 1.43 C is at rest and is located 2.50 cm away from another fix

Question

1)

A charge Q1 = 1.43 C is at rest and is located 2.50 cm away from another fixed charge Q2 = 1.95 C. The first charge is then released. Calculate the kinetic energy of charge Q1 when it is 4.80 cm away from charge Q2.

2)

The potential difference between two parallel conducting plates in vacuum is 200 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 22.0 cm.

Explanation / Answer

Kinetic Energy = Change in potential energy

= kQ1Q2(1/r1 - 1/r2)

= 9*10^9*1.43*10^-6*1.95*10^-6*(1/0.048 - 1/0.025)

= - 0.481 J

Since the amount of potential energy lost, that means the kinetic energy gained by charge.

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