Item 5 Part A A room with 3.0-m-high ceilings has a metal plate on the floor wit

Question

Item 5 Part A A room with 3.0-m-high ceilings has a metal plate on the floor with V OV and a separate metal plate on the ceiling. A 1.2 g glass ball charged to 4.8 nC is shot straight up at 5.3 m/s How high does the ball go if the ceiling voltage is 2.9x106 V? Express your answer to two significant figures and include the appropriate units. h1 = 1.71 Submit My Answers Give Up Incorrect Try Again; 5 attempts remaining Part B How high does the ball go if the ceiling voltage is -3.4x106 V? Express your answer to two significant figures and include the appropriate units. = 2.32

Explanation / Answer

a) Electric field strength between floor and ceiling ..
E = V/d = V/3.0

Downward electric force on +charge .. F = E*q = E*4.80*10^-9 C ..

Grav force on charge = mg = 1.2*10^-3 kg * 9.80 = 11.76*10^-3 N
Net force, Fdown = F + 1.176*10^-2 N
Net acceleration downward = Fdown / m

a = Fdown/1.2*10^-3 kg

At max height (h) final velocity v = 0, initial vel u = 5.3 m/s,
v^2 = u^2 + 2*a*h
0 = 5.3^2 - (2*a*h)

h = 5.3^2/(2*a)

b) Electric field strength between floor and ceiling ..

E = V/d = V/3.0

upward electric force on +charge .. F = E*q = E*4.80*10^-9 C ..

Grav force on charge = mg = 1.2*10^-3 kg * 9.80 = 11.76*10^-3 N

this time the electric force is upward and

net Fup = F - 1.176*10^-2 N

Net acceleration = Fup / m

a = Fup/1.2*10^-3 kg

At max height (h) final velocity v = 0, initial vel u = 5.3 m/s,
v^2 = u^2 + 2*a*h
0 = 5.3^2 - (2*a*h)

h = 5.3^2/(2*a)

Give me ceiling voltage and I will give you final answer solved.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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