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.**

## Expert Answer

**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 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.**