**Is the proton moving towards lower potential or higher potential?****At what potential difference had the proton been stopped?****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, **f****ind the potential difference** if the proton’s **initial speed is 100,000 m/s**.

Plugging values in the **equation (2):**

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

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