Question

A man of 50.0 kg mass is in equilibrium between two parallel plates. The man is charged up to + 130. mC of charge. The plates are 1.75 meters apart.

a. Draw a Free Body Diagram of the man. Hint: He is like a giant oil drop.

b. What must be the value of the electric force to keep him suspended between the plates?

c. What is the value of the electric field between the plates?

d. What must be the obvious direction of the electric force?

e. Is the top plate positive or negative?

f. What is the potential difference between the plates?

g. If the area of one plate is 20.0 m², what is the charge on one of the plates?

2. An electron initially at rest is accelerated between two parallel plates 2.000 cm apart. The area between the plates is in a vacuum. The potential difference across the plates is 300.0 V.

a.What kinetic energy does the electron have when it hits the positive plate?

b.What is the electric field between the plates?

c.What is the electric force on the electron between the plates?

d.If the plates have an area of 400.0mm², what is the capacitance of these plates?

Answer 1

2.

The separation between the plate is (d) =2cm =0.02m

The potential difference across the plates is (V) =300.0 V.

Area of the plates is (A) =400mm2 =400*10^-6m2

The capacitance of the capacitor is given by

             C =eoA/d =8.85*10^-12*400*10^-6/0.02 =177000*10^-18F=0.177pF

The electic field between the plates is given by

             E =V/d =300/0.02 =15000V/m

The electic force between the plates is given by F =qE =(1.6*10^-19)(15000) =24000*10^-19N=2.4*10^-15N

The kinetic energy is given by

                    (1/2)mv2 =qV then v =Sqrt(2qV/m) =Sqrt(2*1.6*10-19*300/9.1*10-31) =Sqrt((960/9.1)*10¹²)=10.27*10⁶m/s

KE =(1/2)mv² =0.5*9.11*10^-31(10.27*10⁶)² =480.42*10^-19J =4.80*10^-17J