In the device below a 500 V difference of potential is maintained between two co

Question

In the device below a 500 V difference of potential is maintained between two conducting surfaces: one is roughly a hemisphere and the other a flat plate. The space between the conductors is a vacuum and the electric field in this space is non uniform. Several equipotential surfaces are shown. An electron crosses the 300 V equipotential surface heading in the general direction of the metal plate with a speed of 1.45 times 10' m/s. a) Determine the electron's kinetic energy as it crosses the 300 V surface. How much kinetic energy will the particle have when it reaches the 100 V surface? Also determine its speed at this point. b) How much kinetic energy will the particle have as it reaches the metal plate (assuming it does)? c) Does the electric field (generally) point in the direction of the flat plate or the hemisphere? How can tell? Discuss.

Explanation / Answer

then,According o to the question

we know that

initial speed,u =1.45*10^7

potential difference of the surface=300V

so,

work done,W=q*V=1.6*10^-19*300=4.8*10^-17J

This work is referred as Kinetic energy, So the kinetic energy of the particle when it crosses 300V surface is

4.8*10^-17J

the kinetic energy of the particle when it crosses 100V surface is

potential difference,dV=100-300=-200V

W=q*V=1.6*10^-19*(-200)=-3.2*10^-17J

the kinetic energy of the particle when it crosses 100V surface is -3.2*10^-17J

speed at this point is,

K.E=(1/2)*m*v2= -3.2*10^-17J

v2=2*(-3.2*10^-17)J/9.1*10^-31=6.8*10^13

so v=8.2*10^6m/then,

b) Kinetic energy when it reaches metal plate is

W=q*V=1.6*10^-19*(500)=8*10^-17J

Kinetic energy when it reaches metal plate is 8*10^-17J

c)

Electric field in the direction point towards positive upward direction in the direction of the plate.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---