Question

A room with 2.7-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.5 g glass ball charged to 5.0 nC is shot straight up at 4.7 m/s .

How high does the ball go if the ceiling voltage is 2.9×10⁶ V ?

How high does the ball go if the ceiling voltage is −2.9×10⁶ V ?

Answer 1

for 2.9*10^6 V

this potential difference creates an electric field in dowward direction.Hence the ball experiences downward electric force along with weight

net force on the ball = weight (m*g) + electric forcec(Fe)

Fnet = m*g + q*E

Electric field is E = V/d = 2.9*10^6/2.7 = 1.07*10^6 V/m

Fnet = (1.5*10^-3*9.81) + (5*10^-9*1.07*10^6)

Fnet = 20.065*10^-3 N

Work done by this net force= change in kinetic energy

Fnet*S = 0.5*m*u^2

required height is S = 0.5*m*u^2/Fnet = (0.5*1.5*10^-3*4.7*4.7)/(20.065*10^-3) = 0.825 m

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For -2.9*10^6 V

this potential difference creates an electric field in upward direction.Hence the ball experiences upward electric force along with down ward weight

net force on the ball = - weight (m*g) + electric forcec(Fe)

Fnet = -m*g + q*E

Electric field is E = V/d = 2.9*10^6/2.7 = 1.07*10^6 V/m

Fnet = -(1.5*10^-3*9.81) + (5*10^-9*1.07*10^6)

Fnet = -9.365*10^-3 N

Work done by this net force= change in kinetic energy

Fnet*S = 0.5*m*u^2

required height is S = 0.5*m*u^2/Fnet = (0.5*1.5*10^-3*4.7*4.7)/(9.365*10^-3) = 1.76 m