How much work energy will it take to move a particle with a

How much work (energy) will it take to move a particle with a charge of +2e particle from G to F? Recall that e = 1.6x10-19 C, the magnitude of charge on an electron or proton. Give your answer in units of electron-Volts (eV). The electron volt is defined in Sec. 16.3 of your book.

The work done by you is positive if you have to push the charge against an electric force, and negative if the electric force is pulling the charge in that direction anyway.

Assuming the electric potential energy of any particle at the bar electrode at right to be zero, what is the electrical potential energy (in eV) of this +2e charged particle when it is at F?

How much work, in eV, will it take to move this +2e charged particle from F to E?

How much work in eV will it take to move this +2e charged particle from E to C?

What is the electrical potential energy in eV of the +2e charged particle at point C?

What is the electric field strength in volts/m at H?

Hint: H is 1 cm from A. You can assume that close to A, the electric fields and potentials are “close enough” to that of a point charge (see Fig. 2) that you can use the description of point charge potentials (see Sec. 16.2) to answer this question and the next 2.

Which particle will feel the force of greatest magnitude at point H: q = +2e or q = –3e?

What force in Newtons would be exerted on the +2e charged particle by the electric field at point H?

What is the electrical potential energy, in eV, of the +2e charged particle at point A?

If the +2e charged particle is released from rest at Point A, what is its kinetic energy in eV when it gets back to the 12 Volt equipotential line?

With what speed, in m/sec, will this +2e charged particle, released from rest at A, crash into the bar electrode? Take the mass of the particle to be M = 6.4 x 10-19 kg.

How much will a object of charge ?3e change its electrical potential energy (include increase or decrease) in going from F to B.

Bar Point Electrode 12 Volts \\18Vults 24 Volts Equipotential Lines

Solution

work done from g to f

w = V q =(6-0) x 2 e =12 e =12 x 1.6 x 10-19 =19.2 x 10^-19 joule

potential enery at F =19.2 x 10^-19 joule

since potential of both point e and f is same so potential difference is zero

so W = 0 x 2e =0 , no work required to move from e to f

from e to c potetial difference is 18-6 =12

than work required = Vq = 12 x 2e =38.4 x 10^-19 joule

electrical potential energy at c is = 18 x 2 e = 57.6 X 10^-19 j

E=V/d

since it is very close to A V approx equal to 30 volt

E = 30/(.01 ) = 3000

1 cm =.01 m

force at point H = F=qE = 2 x 1.6 x 10^-19 x 3000 = 9.6 x 10^-16 Newton

What is the electrical potential energy, in eV, of the +2e charged particle at point A

w = 30 x 2 e = 60e joule

If the +2e charged particle is released from rest at Point A, what is its kinetic energy in eV when it gets back to the 12 Volt equipotential line?

k.e = work done =( 30-12)x2e=18 x 2e=36e joule

With what speed, in m/sec, will this +2e charged particle, released from rest at A, crash into the bar electrode? Take the mass of the particle to be M = 6.4 x 10-19 kg.

1/2 mv2 =Vq

1/2 M V2 = 30 x 2 e

V= sqrt( 4 x 30 e/m) =5.477 m/sec

How much will a object of charge 3e change its electrical potential energy (include increase or decrease) in going from F to B.

W=(24-6)x (-3e)

= -54e joule - sign indicates decrease