**When an electron with an initial speed of $ 6.00 \times 10^5 $ m/s, due to an electric field is brought to rest.**

**Find a region i.e., either a higher potential or lower potential that the electron will move.****Find the potential difference that is required to stop the electron.****Find the initial kinetic energy in electron volts**of the electron.

This question aims to find the region of the electron in which it moves, i.e., either a higher or a lower **potential** when it is moving. Furthermore, the **potential difference** required to stop and the initial **kinetic energy** of the electron is also calculated.

Moreover, this question is based on the concept of an electric field. The electrical potential is the amount of **work** that is needed to move a unit charge from one point to another specific point.

## Expert Answer

a ) From the concept of **potential difference**, we know that the electrons move from higher potential to a lower potential to get in rest.

b ) The stopping potential difference can be calculated as follows:

**mass of electron** = $ m = 9.11 \times 10^{-31} kg $

**charge on electron** = $ e = 1.602 \times 10^{-19}C $

**electron initial speed** = $ v = 6.00 \times 10^5 m/s $

\[ \dfrac{mv^2}{2} = -q \Delta V\]

Therefore, by substituting the above values, we have:

\[ \Delta V = \dfrac{(9.11 \times 10^{- 31} kg) (6.00 \times 10^5 m/s )^2} {2 (1.602 \times 10 ^{- 19}C) } \]

\[ = 102.4 \times 10^{-2} V \]

\[ = 1.02 V \]

c ) The initial** kinetic energy** of the electrons in electron volt is:

\[ \Delta K = \dfrac {m v^2} {2} \]

\[ = \dfrac{(9.11 \times 10^{ -31 } kg) (6.00 \times 10^5 m/s )^2} {2} \]

\[ 1.64 \times 10^ {- 19}J (\dfrac{1eV}{1.602 \times 10 ^{ -19 }C}) \]

\[ = 1.02eV \]

## Numerical Results

\[ \Delta K = 1.02eV \]

## Example:

**work done**in moving a

**charge**of $20 mC$ from infinity to a point O in an electric field is $15 J$, then what is the electric potential at this point?

**Solution:**

Work done = $W = 20 mC$

Charge = $q = 15 J$

Potential difference = $P. D = ?$

and the work done is:

\[ W = \dfrac {P. D}{q} \]

\[ P. D = \dfrac {q}{W} \]

\[ = 15 \times 20 \times 10^{- 3} \]

\[ = 300 \times 10^{- 3} V \]

*Images/ Mathematical drawings are created with Geogebra.*