Question

Consider two points in an electric field. The potential at point 1, V1, is 24V. The potential at point 2, V2, is 154V. A proton is moved from point 1 to point 2.

(a) Write an equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).
(c) Express v2, the speed of the electron at point 2, in terms of AU, and the mass of the electron me.
(d) Find the numerical value of v2 in m/s

Answer 1

The change in electric potential energy (ΔU) of a proton when moved from point 1 to point 2 in an electric field is calculated with the formula ΔU = e * (V2 - V1). The numerical value for ΔU in electron volts (eV) is approximately 130 eV. However, there is an inconsistency in the question when it asks for the speed of an electron rather than a proton.

Step-by-step explanation:

When considering a proton moved from one point to another in an electric field, we can calculate the change in electric potential energy using the equation ΔU = q * ΔV, where ΔU is the change in potential energy, q is the charge of the proton, and ΔV is the change in electric potential. For a proton, this is ΔU = e * (V2 - V1).

(a) The equation for the change in electric potential energy (ΔU) of the proton is ΔU = e * (V2 - V1).

(b) The numerical value of ΔU in electron volts is ΔU = (1.602 × 10^-19 C) * (154V - 24V) = 1.602 × 10^-19 C * 130V = 20.826 × 10^-18 J. Since 1 eV = 1.602 × 10^-19 J, ΔU is approximately 130 eV.

(c) The speed v2 of an electron at point 2 in terms of ΔU and the mass of the electron me can be expressed using the kinetic energy equation. For an electron (not a proton, as initially stated), the kinetic energy K.E. = (1/2) me * v2^2 = ΔU. Solving for v2 gives v2 = √(2 * ΔU / me).

(d) To find the numerical value of v2, we would need the mass of the electron and ΔU. Assuming ΔU is the kinetic energy given to the electron and the mass of the electron me is approximately 9.109 × 10^-31 kg, v2 = √(2 * ΔU / me). Inserting the values gives v2 = √(2 * 20.826 × 10^-18 J / 9.109 × 10^-31 kg) which needs to be computed to obtain a numerical value. However, the initial question contains an inconsistency by referring to a proton and then asking for the speed of an electron. Please clarify.

Answer 2

`\triangle U=-e (V_2-V_1)`

`\triangle U=130eV`

`V_2=\sqrt{ (2)/(me)((1)/(2)meV_1^2+e(V_2-V_1)}`

Step-by-step explanation:

From the question we are told that

The potential at point 1,
`V_1 = 24V`

The potential at point 2,
`V_2 = 154V`

a)Generally work done by proton is given as

`w=-\triangle U`

`e\triangle V=-\triangle U`

`\triangle U=-e (V_2-V_1)`

Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as

`\triangle U=-e (V_2-V_1)`

b)Generally the electric potential energy in electron volts (eV). is mathematically given as

`\triangle U=-e (154-24)V`

`|\triangle U| =|-e (130)V|`

`\triangle U=130eV`

c) Generally according to the law of conservation of energy

`(K.E+P.E)_1=(K.E+P.E)_2`

`(1)/(2)meV_1^2+eV_1 =(1)/(2)mev_2^2+eV_2`

`V_2^2=(2)/(me)((1)/(2)meV_1^2+e(V_2-V_1)`

`V_2=\sqrt{ (2)/(me)((1)/(2)meV_1^2+e(V_2-V_1)}`