A uniform electric field of magnitude 656 N/C exists between two parallel plates

Question

A uniform electric field of magnitude 656 N/C exists between two parallel plates that are 4.40 cm apart. A proton is released from rest at the positive plate at the same instant an electron is released from rest at the negative plate. (a) Determine the distance from the positive plate at which the two pass each other. Ignore the electrical attraction between the proton and electron. (b) Repeat part (a) for a sodium ion (Na+) and a chloride ion (CI) cm Need Help? ReitChat About Need Help? -Read it t About It 12 points SerPSE8 23.P.061. Three point charges are aligned along the x axis as shown in the figure below My Notes Ask Your Teacher 0.500 m 0.800 m -4.00 nC 5.00 nC 3.00 nC Find the electric field at the following positions (a) (3.10, 0) N/C (b) (0, 3.10) N/C

Explanation / Answer

here

F = q *E = m *a

x(t) = x0 + 0.5 * a *t^2 = x0 + 0.5 * q *E / m t^2

for the proton

xp(t) = 0.5 *( q *E / mp ) t^2

t^2 = xp * mp * 2 / q*E

for the electron

xe(t) = d - 0.5 *( q *E / me) * t^2

t^2 = (d - xe) * 2 * me / q*E

at the time they pass xp = xe

xp * mp * 2 / q *E = (d -xp) *2 * me /q*E

xp * mp = d* me - xp *me

xp( mp + me ) = d *me

xp = d * me / ( mp + me)

xp = 0.044 * 9.11 * 10^-31 / ( 1.67 * 10^-27 + 9.11 * 10^-31 )

xp = 2.39 * 10^-5

so the distance form the postive plate is 2.39 * 10^-5 m

b)

here also we use the same formula

xp = d * me / ( mp + me)

here the mass of the sodium ion(+) = 3.81 * 10^-26 kg

mass of chloride = 5.88 * 10^-26 kg

then put in the formula

xp = 0.044 * 5.88 * 10^-26 / 3.81 * 10^-26 + 5.88 * 10^-26

xp = 0.0267 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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