A negatively charged particle is released from rest at point B and accelerates u

Question

A negatively charged particle is released from rest at pointB and accelerates until it reachespoint A. The mass and charge of the particleare 4.0 x 10 ^-6 kg and -2.0 x 10^-5 C, respectively. Only the gravitational force and the electrostatic force act on theparticle, which moves on a horizontal straight line withoutrotating. The electric potential at A is 36 Vgreater than that at B; in other words, Va-Vb=36 V. What is the translational speed of the particle at pointA? I will rate lifesaver right away!! A negatively charged particle is released from rest at pointB and accelerates until it reachespoint A. The mass and charge of the particleare 4.0 x 10 ^-6 kg and -2.0 x 10^-5 C, respectively. Only the gravitational force and the electrostatic force act on theparticle, which moves on a horizontal straight line withoutrotating. The electric potential at A is 36 Vgreater than that at B; in other words, Va-Vb=36 V. What is the translational speed of the particle at pointA? I will rate lifesaver right away!!

Explanation / Answer

conservation of energy     b=a 1/2mv^2+mgh+epe=1/2mv^2+mgh+epe solve for va va=sqrt(vb^2 +((2q(va-vb))/m)) sqrt(2(-2e-5)(-36v)/4e-6kg)) =sqrt(360) =18.97 its simple just set, the traslation speed is close tot he kineticenergy, so energy is conserve so just set them equal to each other,then the rest are just like algebra sove for v

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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