Question

A positive charge of 4.60 C is fixed in place. From a distance of 3.10 cm a particle of mass 6.30 g and charge +3.80 C is fired with an initial speed of 74.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get' before it comes to rest and starts traveling away? Submit Answer Tries 0/20 e Post Discussion Send Feedback

Answer 1

apply energy conservation, since no external forces acting on the system.

Initially,

Potential Energy of two-charge system = kq₁q₂/r

= 9*10^9*4.60*10^-6*3.80*10^-6/ 0.0310 = 5.07483871 J  

Kinetic Energy of particle of mass 6.3 g = 1/2*0.0063*74^2 =17.2494 J

Finally,  

let the particle stop at a distance d from the charge.

Potential Energy of two-charge system = kq₁q₂/d

= 9*10^9*4.60*10^-6*3.80*10^-6/ d

Kinetic Energy of the particle of mass 6.3 g = 0

Eqauting the total intial energy and final energy, we get,

5.07483871 + 17.2494 = 9*10^9*4.60*10^-6*3.80*10^-6/ d

5.07483871 + 17.2494-9 10° 4.6 10-6 :

10.15732 22.3242 =

from here d ≈ 0.00704705 m =7.04705*10^-3 m answer

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