Suppose you wanted to use the electrostatic force to levitate a massive object a

Question

Suppose you wanted to use the electrostatic force to levitate a massive object against gravity. In principle, this could be accomplished by introducing enough excess charge on two conducting surfaces, as long as the charges had the same sign (see figure below). If the mass of the payload is 1000 kg, what is the magnitude electric force required to suspend it against gravity? If the payload has a total charge of -10 C, how much excess charge is required on the rail to levitate the payload against gravity? The rail and payload are separated by a distance of 0.01 m.

Explanation / Answer

a) electric force must be equal to gravitational force but in opposite direction.

Fe = Fg

= m*g

= 1000*9.8

= 9800 N

b) Let Q2 is the excess charge needed.

Apply, Fe = k*Q1*Q2/d^2

Q2 = Fe*d^2/(k*Q1)

= 9800*0.01^2/(9*10^-9*-10)

= -1.09*10^7 C

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