A positive charge of 4.60 µC is fixed in place. From a distance of 4.60 cm a par

Question

A positive charge of 4.60 µC is fixed in place. From a distance of 4.60 cm a particle of mass 5.70 g and charge +3.70 µC is fired with an initial speed of 68.0 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?

(The particle comes to rest when it has no kinetic energy. The change in kinetic energy will be equal to the increase in electrical potential energy. Note that the particle has some potential energy at the starting point. )

Explanation / Answer

Given a positive charge of 4.60µC is fixed in place.Froma distance of 4.60cm a particle of mass 5.70g and charge +3.70µC is fired with an initial speed of 68.0m/s directlytoward the fixed charge. When a charge is fired from a distance and it comes close to theother charge then it stops at a distance equal to the distance ofclosest approach and at this point the potential energy of thecharge being fired becomes equal to its kinetic energy.Therefore,weget (1/2)mv2 = (1/4peo) x(q1q2/d) -----------(1) where m is the mass of the particle being fired,v is itsvelocity,q1 its charge and q2 the chargetowards which it is approaching. Substituting the values of m,v,q1 andq2 in the equation (1),we get d = (1/4peo) xq1q2/(1/2)mv2 =(2/4peo) x(q1q2/mv2) = ... Therefore the charge + 3.70µC stops at a distance of... from the charge 4.60µC and starts travellingaway.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.