
How much is the charge of an electron?
Answer
568.8k+ views
Hint: We know that electrons are responsible for the flow of current. Also when electrons are charged they allow the current to flow through them. So, we need to relate current, electrons and charge in order to find the solution of the given question.
Complete step by step solution:
We know that, Current = Net flow of charge / per unit time
i.e. $I = \dfrac{Q}{t}$
Also if we take a look on the units, it can be written as
$1A = \dfrac{{1C}}{{1s}}$ ……………………(i)
The unit of current is Ampere, A. For charge it is Coulomb, C and for time is second, s.
We also know that, 1 Coulomb of charge contains $6 \times {10^{18}}$ electrons
Now, we need to put the value of 1 coulomb in equation (i)
$\Rightarrow 1A = \dfrac{{6 \times {{10}^{18}}}}{1}$$\dfrac{{electrons}}{{\sec }}$
$\Rightarrow 6 \times {10^{18}}electrons = 1A \times 1s$
$\Rightarrow$ $1 electron$ = $\dfrac{{1 \times 1}}{{6 \times {{10}^{18}}}}$$A \times s$
$\Rightarrow$ $1 electron$ = $1.6 \times {10^{ - 19}}$$A \times s$
Also from equation (i) we know that $1C = 1A \times 1s$
Therefore, $1 electron$ = $1.6 \times {10^{ - 19}}C$
Hence, we can conclude that one electron contains $1.6 \times {10^{ - 19}}C$ of charge.
Note: We should remember that the smallest unit in which charge can be measured is in the terms of electron count. Also, charges follow the quantization rule. According to which $Q = ne$, which indicates that charges can be only multiplied with integers (n).
Complete step by step solution:
We know that, Current = Net flow of charge / per unit time
i.e. $I = \dfrac{Q}{t}$
Also if we take a look on the units, it can be written as
$1A = \dfrac{{1C}}{{1s}}$ ……………………(i)
The unit of current is Ampere, A. For charge it is Coulomb, C and for time is second, s.
We also know that, 1 Coulomb of charge contains $6 \times {10^{18}}$ electrons
Now, we need to put the value of 1 coulomb in equation (i)
$\Rightarrow 1A = \dfrac{{6 \times {{10}^{18}}}}{1}$$\dfrac{{electrons}}{{\sec }}$
$\Rightarrow 6 \times {10^{18}}electrons = 1A \times 1s$
$\Rightarrow$ $1 electron$ = $\dfrac{{1 \times 1}}{{6 \times {{10}^{18}}}}$$A \times s$
$\Rightarrow$ $1 electron$ = $1.6 \times {10^{ - 19}}$$A \times s$
Also from equation (i) we know that $1C = 1A \times 1s$
Therefore, $1 electron$ = $1.6 \times {10^{ - 19}}C$
Hence, we can conclude that one electron contains $1.6 \times {10^{ - 19}}C$ of charge.
Note: We should remember that the smallest unit in which charge can be measured is in the terms of electron count. Also, charges follow the quantization rule. According to which $Q = ne$, which indicates that charges can be only multiplied with integers (n).
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