Enhanced with Solution A small particle has charge -3.80 mu C and mass 1.40 time

Question

Enhanced with Solution A small particle has charge -3.80 mu C and mass 1.40 times 10^-4 kg. It moves from point A, where the electric potential is V_A = 300 V, to point B, where the electric potential V_B = 840 Vis greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 5.90 m/s at point A. You may want to review For related problem solving tips and strategies, you may want to view a Video Tutor Solution of Electric force and electric potential. What is its speed at point B? v=_____ Is it moving faster or slower at B than at A? Faster Slower

Explanation / Answer

we know that work done in the electric field is given by change in electric potential= change in energy*electricfield/charge

now change in energy = 1/2mv2^2-1/2mv1^2 where v1=5.9m/s and m=1.4*10^-4kg

charge=q=-3.80microC=-3.80*10^-6C

so potential difference = VB-VA=840-300=540

using the formula 840-300=540=(1/2mv2^2-1/2mv1^2 )/q

or 540=1/2*1.4*10^-4(v2^2-5.9^2)/-3.80*10^-6

or 540*-3.80*10^-6 *2/1.4*10^-4=(v2^2-5.9^2)

or v2=2.345m/s

So it is moving slower than A

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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