contestada

Consider two points in an electric field. The potential at point 1,
V
1
is 31 V.

The potential at point 2,
V
2
is 168 V. A proton is moved from point 1 to point 2.

a) Write an equation for the change of the electric potential energy
Δ
U of the proton, in terms of the symbol given and the charge of the proton e.

b) Solve for the numerical value of the change of the electric potential energy in electron volts (eV).

c) How much work is done by the electric force during the motion of the proton in J?

Respuesta :

Answer:

(a) [tex]\Delta U=(V_2-V_1)e[/tex]

(b) 137 eV

(c) [tex]2.192\times10^{-17}J[/tex]

Explanation:

Since electric field is conservative, the change in electric potential energy is simply the difference between the potential energy at position 2 and 1:

[tex]\Delta U=(V_2-V_1)e=(168-31)\times 1 eV=137eV[/tex]

1 electron volt is defined as the energy required to move an electron (proton) and accelerating it through a potential difference of 1 volt.

The work done by the electric force in units of J is given by,

[tex]\Delta U=137\times(1.6\times10^{-19})J=2.192\times10^{-17}J[/tex]

ajeigbeibraheem

Considering two points in an electric field. The equation for the change of the electric potential energy is  [tex]\mathbf{\Delta U =U_2-U_1}[/tex] and the amount of work done by the electric force is -2.2742 × 10⁻¹⁷ J

The electric potential energy at any given point is the amount of potential energy required to push a unit charge from a given point to another against an electric field.

From the given information;

  • The electrical potential energy at Point P₁ is determined by using the relation;

[tex]\mathbf{U_1 = \dfrac{kqQ}{r}}[/tex]

where;

  • U₁ = change in electrical potential energy
  • k = constant
  • Q = charge
  • r = distance from the point charge

[tex]\mathbf{U_1 = V_1q} \ \ \ \ \mathbf{ since \ V = \dfrac{kQ}{r}}\ \\ \\ \\ \mathbf{U_1 = 31q}[/tex]

The electrical potential energy at point P₂ is:

[tex]\mathbf{U_2= V_2q} \ \ \ \ \\ \\ \\ \mathbf{U_2 = 168 \times q}[/tex]

∴

The equation for the change in the electric potential energy is [tex]\mathbf{\Delta U =U_2-U_1}[/tex]

(b)

Solving the equation, we have:

ΔU = 168 - 31

ΔU = 137 eV

(c)

The work done by the electric force during the motion of the proton is negative since the charge in the change in the potential difference is positive.

i.e.

[tex]\mathbf{\Delta U_E= q(137) }[/tex]

[tex]\mathbf{\Delta U_E= (-e) (137) \ V}[/tex]

[tex]\mathbf{\Delta U_E= -137 \ eV}[/tex]

where;

  • [tex]\mathbf{\Delta U_E}[/tex] denotes the work done by the electric force.

Recall that:

  • 1 electron volt = 1.66 × 10 ⁻¹⁹ Joules

∴

  • [tex]\mathbf{\Delta U_E= -137 \times 1.66 \times 10^{-19} \ J}[/tex]
  • [tex]\mathbf{\Delta U_E= -2.2742 \times 10^{-17} \ J}[/tex]

Therefore, we can conclude that the equation for the change of the electric potential energy is  [tex]\mathbf{\Delta U =U_2-U_1}[/tex] and the amount of work done by the electric force is -2.2742 × 10⁻¹⁷ J

Learn more about potential energy here:

https://brainly.com/question/17858145?referrer=searchResults

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