Lippert wants to levitate a frog of mass 0. 2 kg. The frog is sitting on a flat,

Question

Lippert wants to levitate a frog of mass 0. 2 kg. The frog is sitting on a flat, horizontal table, and Prof. Lippert arranges a positively charged plate 0. 4 m directly above the frog, parallel to the table, which generates a constant vertical electric field of 5 N/C pointing straight downward. The frog is then given an electrical charge. (a) What is the charge needed in order for the frog to begin rising up off the table? (b) For each of the three forces acting on the frog, the Coulomb force from the charged plate, the gravitational force from the earth, and the normal force from the table, describe the reaction force which is a consequence of Newton's Third Law. In cach ease, explain on what object and in which direction the reaction force acts. (c) Suppose the frog is now 0. 2 m above the table and has a charge of -0. 3 C. What is the frog's acceleration? Does it rise or fall?

Explanation / Answer

I am sure Prof. Lippert is a very nice guy. Though...

a) The condition to start levitation: Fe1=Fg <--> qE1=mg --> q=mg/E1=-0.392 C. It should be negative in order to be attracted by the upper plate.

b) If the frog is sitting on the plate: N+Fe=Fg --> N=Fg-Fe

Fe=from the frog to the upper plate (attraction force of the upper plate)

N=from the plate (on which the frog is sitting) upward (reaction of the plate to the weight of the frog)

Fg=from the frog to the ground (attraction of the Earth)

c) The charge of the upper plate is involved in the eq of E: E1=kQ/r12 (at the distance of 0.4 m)

E2=kQ/r22 (at the distance of 0.2 m).

Divide these 2 eq, work on the new eq and get: Q=E1r12/k=E2r22/k --> E2=E1r12/r22 (1)

k=Coulomb's constant (9*109Nm2/C2)

Num: E2=20V/m

At 0.2 m we have (consider Oy axis as usual): Fe2-mg=ma --> a=(qE2-mg)/m=20.2 m/s2

It seems it flies, and quite rapidly. Nice cartoon...

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