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A proton with an initial speed of 650,000 m/s is brought to rest by an electric field.

  1. Is the proton moving towards lower potential or higher potential?
  2. At what potential difference had the proton been stopped?
  3. How much kinetic energy (in electron-volts) did the proton carry at the start of the journey?

The aim of this question is to understand the interaction of charged bodies with electric fields in terms of kinetic energy and potential energy.

Here we will use the concept of potential gradient, which is mathematically described as:

\[ PE \ = \ \dfrac{ U }{ q } \]

Where PE is the potential energy, U is the electric potential and q is the charge.

The kinetic energy of any moving object is defined mathematically as:

\[ KE \ = \ \dfrac{ mv^2 }{ 2 } \]

Where m is the mass of the moving object and v is the speed.

Expert Answer

Part (a) – Since the proton is positively charged and gradually decelerates to rest, it must be moving towards a higher potential region.

Part (b) – From law of conservation of energy:

\[ KE_i \ + \ PE_i \ = \ KE_f \ + \ PE_f \ … \ … \ … \ (1) \]

where KE and PE are the kinetic and potential energies, respectively.

Since:

\[ PE \ = \ \dfrac{ U }{ q } \]

and:

\[ KE \ = \ \dfrac{ mv^2 }{ 2 } \]

Equation (1) becomes:

\[ \dfrac{ mv_i^2 }{ 2 } \ + \ \dfrac{ U_i }{ q } \ = \ \dfrac{ mv_f^2 }{ 2 } \ + \ \dfrac{ U_f }{ q } \]

Rearranging:

\[ U_f \ – \ U_i \ = \ \dfrac{ \frac{ m }{ 2 } ( \ v_i^2 \ – \ v_f^2 \ ) }{ q } \ … \ … \ … \ (2) \]

Given that:

\[ v_i \ = \ 650000 \ m/s \]

\[ v_f \ = \ 0 \ m/s \]

For proton, we know that:

\[ m \ = \ 1.673 \ \times \ 10^{ -27 } \ kg \]

And:

\[ q \ = \ 1.602 \ \times \ 10^{ -19 } \ C \]

Plugging these values in the equation (2):

\[ U_f \ – \ U_i \ = \ \dfrac{ \dfrac{ 1.673 \ \times \ 10^{ -27 } }{ 2 } ( \ 650000^2 \ – \ 0^2 \ ) }{ 1.602 \ \times \ 10^{ -19 } } \]

\[ \Rightarrow U_f \ – \ U_i \ = \ 2206.12 \ Volt \]

Part (c)Initial kinetic energy is given by:

\[ KE_i \ = \ \dfrac{ mv_i^2 }{ 2 } \]

\[ KE_i \ = \ \dfrac{ (1.673 \ \times \ 10^{ -27 } ) (650000)^2 }{ 2 } \]

\[ KE_i \ = \ 3.53 \times 10^{ -16 } \ J\]

Since $ 1J \ = \ 6.24 \times 10^{ 18 } \ eV $:

\[ KE_i \ = \ 3.53 \times 10^{ -16 } \times 6.24 \times 10^{ 18 } \ eV\]

\[ \Rightarrow KE_i \ = \ 2206.12 \ eV\]

Numerical Result

Part (a): Proton moves towards higher potential region.

Part (b): $ U_f \ – \ U_i \ = \ 2206.12 \ V $

Part (c): $ KE_i \ = \ 2206.12 \ eV $

Example

In the same scenario given above, find the potential difference if the proton’s initial speed is 100,000 m/s.

Plugging values in the equation (2):

\[ U_f \ – \ U_i \ = \ \dfrac{ \dfrac{ 1.673 \ \times \ 10^{ -27 } }{ 2 } ( \ 100000^2 \ – \ 0^2 \ ) }{ 1.602 \ \times \ 10^{ -19 } } \]

\[ \Rightarrow U_f \ – \ U_i \ = \ 52.21 \ Volt \]

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