What is the kinetic energy of a particle of mass 6.8 times 10^-27 kg and charge

Question

What is the kinetic energy of a particle of mass 6.8 times 10^-27 kg and charge +3.2 times 10^-19 C, moving with a velocity of 1.0 times 10^7 m/s. What is the distance of closest approach if potential energy of a Q_2 11.2 times 10^19 C was able to stop the approaching particle? Explain the following using equations, words and sketches: Force between charges, Electric field, potential, and potential energy. In the figure below C_2 (3uF) and C_2 (6uF) are in series across 90 V d.c. supply. Calculate the Charges on C_1 and C_2 and potential difference across each and the energy stored in each. Suppose they are connected in parallel as shown find the charges in each capacitor.

Explanation / Answer

kinetic energy = 0.5 * mv^2

kinetic energy = 0.5 * 6.8 * 10^-27 * (1 * 10^7)^2

kinetic energy = 3.4 * 10^-13 J

by the conservation of energy

initial energy = final energy

3.4 * 10^-13 = k * q1 * q2 / r

3.4 * 10^-13 = 9 * 10^9 * 3.2 * 10^-19 * 11.2 * 10^-19 / r

r = 9.4870588 * 10^-15 m

closest distance it'll approach = 9.4870588 * 10^-15 m

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