Question

Three protons occupy the vertex of an equilateral triangle with each side of 1 cm. Mp= 1.67E-27 kg, qp= 1.6E-19 C.

a) release all three protons from rest, what is the max speed of each proton?

b) Only one proton is released while the other two are fixed. Find the speed of the moving proton after a long time.

Answer 1

a)

U_i = initial total electric potential energy of the system = 3 k q_p²/d

U_f = final total electric potential energy of the system = 0 J

KE_f = final kinetic energy of three protons

KE_i = initial total kinetic energy of three protons = 0 J

using conservation of energy

U_i + KE_i = U_f + KE_f ​​​​​​​

3 k q_p²/d + 0 = 0 + KE_f ​​​​​​​

3 k q_p²/d = 3 (0.5) m_p v²

k q_p²/d = (0.5) m_p v²

(9 x 10⁹) (1.6 x 10^-19)²/0.01 = (0.5) (1.67 x 10^-27) v²

v = 5.25 m/s

b)

d = distance between each pair of charges = length of side of triangle = 1 cm = 0.01 m

q_p = charge at each vertex = 1.6 x 10^-19 C

m_p = mass of each proton = 1.67 x 10^-27 kg

v = maximum speed of proton = ?

U_i = initial total electric potential energy of a proton = 2 k q_p²/d

using conservation of energy

maximum kinetic energy = initial total electric potential energy of a proton

(0.5) m_p v² = U

(0.5) m_p v² = 2 k q_p²/d

(0.5) (1.67 x 10^-27) v² = 2 (9 x 10⁹) (1.6 x 10^-19)²/0.01

v = 7.43 m/s

b)