What is the elecic pontialngyU of he pair of charges (Take U7 so be ze when the

Question

What is the elecic pontialngyU of he pair of charges (Take U7 so be ze when the dharges hev nseparation AchargeQ.. + 500,Cishald tied at the origin Asecond point charge q 00 Cwith mass of 2 90 10g is placed an the -axis 0 250 m from the + Part The second point charge seleasemrest What is its speed when its diotance from the origin is 500m Submi Part C The second pont charge is released tom reat What is t seed when its stanceom the origin is 5 00m Submit Part D The second point charge is released tom res What s ts apeed whent stance fom the origin is 500

Explanation / Answer

Given,

First charge Q= + 5 uC

second point charge q= +2 uC

mass m = 2.90×10-4 kg

distance, r = 0.250 m

a) The electric potential energy, U1= kQq/r1 =(9*10^9)[5*10^-6][2*10^-6] / 0.250 = 0.36 J

b) The electric potential energy=U2=kQq/r2 =(9*10^9)[5*10^-6][2*10^-6] / 0.5 =0.18 J

Kinetic energy = U1 - U2 = 0.18 J

Velocity = v1 = sq rt 2*kinetic energy / mass = sqrt (0.36/2.9*10^-4) = 35.23 m/s

c) The electric potential energy, U3= kQq/r3 =(9*19^9)[5*10^-6][2*10^-6] / 5 = 0.018 J

Kinetic energy = U1 - U3= 0.342 J

Velocity = v2 = sq rt 2*kinetic energy / mass = sqrt (2* 0.342)/(2.9*10^-4) = 48.57 m/s

d) The electric potential energy, U4= kQq/r4 =(9*19^9)[5*10^-6][2*10^-6]/ 50 =0.0018 J

Kinetic energy = U1 - U4 = 0.358 J

Velocity = v3 = sq rt 2*kinetic energy / mass = sqrt (2* 0.358/2.9*10^-4) = 49.70 m/s

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