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Physics

Work in an electric field

1. Characterize the physical quantities: work in an electric field, electric potential and electric voltage.

Solution:

A)
Work in an electric field is the work of external forces when moving a charge in the electric field.

W=E⋅Q0⋅dW = E \cdot Q_0 \cdot d

E=E = intensity of the electric field, Q0=Q_0 = moved charge,
d=d = distance the charge is moved

[W]=1J[W] = 1J

B)
The electric potential at a given point in the electric field is the ratio of the work performed by the electric forces of the electric field when moving a particle with charge Q0Q_0 from that point to the point of zero potential and the charge Q0Q_0.

φ=WQ0=EyQ=k⋅Qr=E⋅d\varphi = \frac{W}{Q_0} = \frac{E_y}{Q} = \frac{k \cdot Q}{r} = E \cdot d [φ]=1J⋅C−1=1V=volt[\varphi] = 1J \cdot C^{-1} = 1V = volt

C)
The electric voltage between two different points of the electric field is equal to the absolute value of the difference in potentials between these points of the electric field.

U=∣φ1−φ2∣=WQ0=E⋅dU = |\varphi_1 - \varphi_2| = \frac{W}{Q_0} = E \cdot d [U]=1V[U] = 1V

 

2.What amount of work does an electric field of intensity 104N.C-1 perform when it moves a charge of 20 μC along the field line for a distance of 10 cm.

Solution:

Analysis:

E = 104N.C-1,  Q0 = 20 μC  = 20.10-6C,  d = 10 cm = 10-1m

fyzika-praca-v-elektrickom-poli-2.gif 

The electric field does work W = 0,02 J

3.What is the potential of a conductor if transferring a charge of 50 μC from a point of zero potential to its surface required work of 0.2 J?

Solution:

Analysis:

Q0 = 50.10-6C, W = 2.10-1J,  φ = ?

fyzika-praca-v-elektrickom-poli-3.gif 

The conductor has potential φ = 4000 V.

4.In a homogeneous electric field with intensity 1 kV.m-1, a particle with charge 25.10-9C moves along a path of 2 m.

  • a) What work do the forces of the electric field perform when moving the particle?
  • b) What is the electric voltage between the starting and ending points of the displacement?

Solution:

Analysis:

E = 103 V.m-1,  Q0 = 25.10-9C,  d = 2 m ,  W = ?  U = ?

fyzika-praca-v-elektrickom-poli-4.gif 

  • The field forces perform work W = 5.10-5J.
  • The voltage between the starting and ending points of the displacement is U = 2 kV.

5.What speed will an electron (Qe= 1,602.10-19C, me = 9,1.10-31kg) reach when passing through a potential difference of 100 V?

Solution:

Analysis:

Qe= 1,602.10-19C, me = 9,1.10-31kg,  Δφ = U = 100 V = 102V

fyzika-praca-v-elektrickom-poli-5.gif 

The electron reaches speed v = 6.106m.s-1.

6.Determine the intensity of the electric field between two parallel conducting plates separated by 5 cm if the voltage between them is 150 V. What work do the field forces perform when transferring a charge of 1 μC from one plate to the other?

Solution:

Analysis:

d = 5cm = 5.10-2m,  U = 150V,  Q0 = 1μC = 10-6C,  E = ?,  W = ?

fyzika-praca-v-elektrickom-poli-6.gif 

The electric field intensity between the two parallel conducting plates is E = 3 kV.m-1. The field forces perform work W = 1,5.10-4J.

7.An alpha particle ( mα = 6,7.10-27kg, Q0α = 3,2.10-19C) entered a homogeneous electric field with velocity 2.106m.s-1. The particle stopped after travelling 2 m.

  • What potential difference did the particle traverse?
  • What is the magnitude of the electric field intensity?

Solution:

Analysis:

mα = 6,7.10-27kg, Q0α = 3,2.10-19C,  v = 2.106m.s-1,  d = 2m,  U = ?, E = ?

fyzika-praca-v-elektrickom-poli-7.gif 

  • The particle traversed potential difference Δφ = U = 4,19.104V.
  • The electric field intensity is E = 2,1.104V.m-1.

8.What electric charge does a microscopic oil droplet of mass 6,4.10-16kg have if it hovers between the plates of a charged capacitor? The capacitor plates are 1 cm apart and the voltage between them is 400 V. (Experiment R.A. Millikan)

Solution:

Analysis:

m = 6,4.10-16kg,  d = 1cm = 10-2m,  U = 4.102V,  g = 10m.s-2,  Q = ?

fyzika-praca-v-elektrickom-poli-8.gif 

The electric charge of the oil droplet is the elementary charge Q = 1,6.10-19C.

9.What work do the forces of the electric field created by charge Q = 2 μC perform when moving charge Q0 = 1 nC from a point 10 cm from charge Q to a point 20 cm from charge Q.

Solution:

Analysis:

Q = 2μC = 2.10-6C,  Q0 = 1 nC = 10.10-9C,  r1 = 0,1m,  r2 = 0,2m, k = 9.109N.m2.C-2

fyzika-praca-v-elektrickom-poli-9.gif 

The electric field forces perform work W = 90 μJ.

10.An electron (e = 1,602.10-19C) moves in a homogeneous electric field from a potential level of 200 V to a level of 300 V. The initial speed is zero, the final speed is 5,93.106m.s-1. Determine:

  • the increase in the electron's energy as it passes between the two potential levels
  • the mass of the electron

Solution:

Analysis:

e = 1,602.10-19C,  U = Δφ = 100V,  v0 = 0,  v = 5,93.106m.s-1

fyzika-praca-v-elektrickom-poli-10.gif 

  • The increase in the electron's energy is ΔW = 1,602.10-17J.
  • The mass of the electron is me = 9,1.10-31kg.