**Calculate the number of electrons the honeybee loses while it develops the given charge while it flies.**

The aim of this article is to find the number of **electrons** being lost by the honeybee while it acquires a **positive charge of +16pC** as it flies through the air.

The basic concept behind this article is the **Electric Charge** and how it is transferred following the **principles of conservation of electric charges**.

**Electric charge** is the charge possessed by **subatomic particles** like **protons, electrons, and neutrons**. **Protons** caries **positive** **electric charge** whereas **negative electric charge** is carried by **electrons**. **Neutrons** are **neutral** and do not carry any electric charge.

**Electric Charge** is represented by the symbol $Q$ or $q$ and the **total electric charge** that is present in a body is equal to the **number of electrons** that the body carries multiplied by the **standard electric charge of an electron** as represented by the following formula:

\[Q\ =\ n\ . e\]

Where:

**Q = Electric charge on the body**

**n = Number of electrons**

**e = Electric Charge on an electron**

The **SI unit** for **Electric Charge** acquired by a body is **Coulomb,** which is represented by **C**.

As a standard, the **electric charge** on an **electron** is $1.6\times{10}^{-19}$

## Expert Answer

Given that:

**Electric Charge on the Honeybee** $Q\ =\ +16pC\ =\ +16\times{10}^{-12}\ C$

**The number of Electrons** $n=?$

When a honeybee flies, it has acquired a **positive charge** but at the same time, it **loses a negative charge** in terms of an **electron** as per the **principles of conservation of electric charge** which states that an **electric charge** can **neither be created nor be destroyed** but it is **transferred** from one system to the other. So the n**et total charge of that system remains the same**.

We know that the **total charge** that has been developed by the honeybee can be represented as follows

\[Q=n\ . e\]

By substituting the values of $Q$ and $e$ in the above expression, we get:

\[16\ \times\ {10}^{-12}\ C\ =\ n\ \times\ (1.6\ \times\ {10}^{-19}\ C) \]

By rearranging the equation:

\[n\ =\ \frac{16\ \times\ {10}^{-12}\ C\ }{1.6\ \times\ {10}^{-19}\ C} \]

\[n\ =\ 10\ \times\ {10}^{-12}\ \times\ {10}^{19}\]

\[n\ =\ 10\ \times\ {10}^{-12}\ \times\ {10}^{19}\]

\[n\ =\ {10}^8\]

The** number of Electrons** is $n\ =\ {10}^8$

## Numerical Result

The **number of electrons** the **honeybee loses** while it develops the given charge while it flies is as follows:

\[n\ =\ {10}^8\]

## Example

When a **plastic ball** is thrown in the air it develops a **charge** of +20pC. Calculate the **number of electrons** the **plastic ball loses** while it develops the given charge while it moves in the air.

Given that:

**Electric Charge on the Plastic Ball** $Q\ =\ +\ 20\ pC\ =\ +\ 20\ \times\ {10}^{-12}\ C$

As we know:

\[Q=n\ . e\]

So:

\[20\ \times\ {10}^{-12}\ C\ =\ n\ \times\ (1.6\ \times\ {10}^{-19}\ C)\]

\[n\ =\ \frac{20\ \times\ {10}^{-12}\ C\ }{1.6\ \times\ {10}^{-19}\ C}\]

\[n\ =\ 12.5\times{10}^7\]

The **number of electrons lost by Plastic Ball** is:

\[n\ =\ 12.5\times{10}^7\]