# A plastic rod is charged to -60 nC by rubbing. (a) Were the electrons added or protons removed from the surface? Explain. (b) How many unit charges (electrons/protons) have been added?

This question aims at developing the understanding of electrostatics, especially the elementary concept of charge transfer through rubbing.

Electrostatics is the branch of physics that deals with the study of charges at rest.

All bodies under a steady state or non-excited state are neutral due to the fact that the number of electrons and protons in them are equal. Whenever two bodies are rubbed together, they exchange electrons. This affects the charge balance and one of the body acquires electrons while the other one loses them.

As a result of this exchange, the body that receives electrons acquires a net negative charge while the one that loses electrons acquires a net positive charge.

The number of electrons lost or gained by a body can be calculated using the following formula:

$\text{ Number of electrons exchanged } = \ \dfrac{ Q }{ e }$

Where Q is the total charge acquired by the body and e is the charge of a single electron which is equal to $1.02 \times 10^{ -27 } \ C$. If the result has a negative sign then it shows that the electrons were lost.

Part (a) – Since the charge on the body is negative, this means that there is an excess of electrons in it. To create such an excess the electrons must have been acquired during the process of rubbing. There is no transfer of protons is possible here because the protons reside inside the nucleus and the strong nuclear force is strong enough to hold them against minute rubbing effects. Hence we can conclude that the electrons were added to the plastic rod.

Part (b) – Calculating the number of electrons acquired by the plastic rod:

$\text{ Number of electrons exchanged } = \ \dfrac{ Q }{ e }$

Given:

$Q \ = \ -60 \ nC \ = \ -60 \ \times \ 10^{ -9 } \ C$

$e \ = \ -1.602 \ \times \ 10^{ -27 } \ C$

So:

$\text{ Number of electrons exchanged } = \ \dfrac{ -60 \ \times \ 10^{ -9 } \ C }{ -1.602 \ \times \ 10^{ -27 } \ C }$

$\Rightarrow \text{ Number of electrons exchanged } = \ 37.45 \ \times \ 10^{ 18 }$

## Numerical Result

Part (a) – Electrons were added to the plastic rod.

Part (b) – Number of electrons added = $37.45 \ \times \ 10^{ 18 }$.

## Example

How many electrons were exchanged if a body acquires a charge of 1 nC?

Using the formula:

$\text{ Number of electrons exchanged } = \ \dfrac{ Q }{ e }$

$\Rightarrow \text{ Number of electrons exchanged } = \ \dfrac{ 1 \ \times \ 10^{ -9 } \ C }{ -1.602 \ \times \ 10^{ -27 } \ C }$

$\Rightarrow \text{ Number of electrons exchanged } = \ -62.42 \ \times \ 10^{ 16 }$

The negative sign shows that the electrons were lost.

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