# Did the electron move into a region of higher potential or lower potential?

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.

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

The potential difference that stopped the electron is:

$\Delta V = 1.02V$

In electron volts, the required initial kinetic energy of the electron is:

$\Delta K = 1.02eV$

## Example:

In a given field, if 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:

The solution can be found as follows:

Work done = $W = 20 mC$
Charge = $q = 15 J$
Potential difference = $P. D = ?$

and the work done is:

$W = \dfrac {P. D}{q}$

$P. D = \dfrac {q}{W}$

$= 15 \times 20 \times 10^{- 3}$

$= 300 \times 10^{- 3} V$

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