A proton with an initial velocity of 2.5 x 10^5 m/s enters at A a region of unif

Question

A proton with an initial velocity of 2.5 x 10^5 m/s enters at A a region of uniform electric field E = 1000 N/c extending over a distance of 30cm from A to B.

What is the kinetic energy of the proton at B?

After reaching B, if the protoin heads toward C, 15 cm straight down from B, what would be its kinetic energy at C?

Answers are 4.0 x 10^-18 J and 4.0 x 10^-18 J. How do I arrive at those answers?

A proton with an initial velocity of 2.5 x 10^5 m/s enters at A a region of uniform electric field E = 1000 N/c extending over a distance of 30cm from A to B. What is the kinetic energy of the proton at B? After reaching B, if the protoin heads toward C, 15 cm straight down from B, what would be its kinetic energy at C? Answers are 4.0 x 10^-18 J and 4.0 x 10^-18 J. How do I arrive at those answers?

Explanation / Answer

Potential due to this electric field V =E.d

Va -Vb = 1000*30/100 = 300 V

So net energy at B = 1/2 mv^2 + eV = 1/2*1.67*10-27 * 6.25*1010 + 1.6*10^(-19) *300 = 4*10^(-18) J (ans)

While moving from B to C ,proton moves perpendicular to E so no change in V from B to C

Vb - Vc =0

So net energy = energy at B = 4*10^(-18) J (ans)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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