roblem Set 1 Begin Date: 1/22/2018 11:00:00 AM -- Due Date: 1/31/2018 10:00:00 A

Question

roblem Set 1 Begin Date: 1/22/2018 11:00:00 AM -- Due Date: 1/31/2018 10:00:00 AM End Date: 2/7/2018 10:00:00 AM (996) Problem 9: A simple and common technique for accelerating electrons is shown in the figure, which depicts a uniform electric field between two plates Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through. Randomized Variables E= 2.25 × 10" N/C ©theexpertta.com 104 NC. Express your answer in meters per Calculate the horizontal component of the electron's acceleration if the field strength is 2.25 second squared, and assume the electric field is pointing in the negative x-direction as shown in the figure Grade Summarv Deductions Potential 0% 100% Submissions Attempts remaining: 9 (190 per attempt) detailed view sinO tan() cos cotan0asin acos atan acotan sinh0 coshOtanhcotanh0 0 END Degrees Radians Submit Hint I give up Hints: 006 deduction per hint. Hints remaining: 2. Feedback: 0% deduction per feedback. All content © 2013 Expert TA, L

Explanation / Answer

Solution:

Let us go to the basics first.

We know that force on the charge (electron in this case) is given by:

F = qE...................Eqn.1 [q = 1.6*10-19 C = charge on electron]

Also, from Newton's equation:

F = ma......................Eqn.2 [m= mass of charge (electron) = 9.109*10-31 kg; a = acceleration]

Thus, eqn.1 becomes:

ma = qE

=> a = qE / m

=> a = (1.6*10-19) * (2.25*104) / (9.109*10-31) = 3.952*1015 m/s2 (Answer)

Thanks!!!

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

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