**If the spheres repel each other with a repulsive force having a magnitude of 3.33X10^(-21) N, calculate the excess electrons that each sphere carries.**

This question aims to find the **number of excess electrons** present on a set of bodies that causes them to **repel each other**.

The basic concept behind this article is the **Electrostatic Force** and **Coulomb’s Law for Charged Bodies**.

The **Electrostatic Force** is defined as one of the fundamental forces in nature that exist between two bodies that carry an **electric charge** and are separated by a **finite distance**. This force can be **repulsive** or **attractive** and varies as the distance between the body changes.

If the **charge** on bodies is **opposite** to each other, the **electrostatic force** is **attractive**. If the **charges** are the** same**, the **electrostatic force is repulsive**.

Its standard unit of measurement is **Newton **$N$.

The **Electrostatic Force** is calculated with the help of **Coulomb’s Law,** which states that the **electrostatic force** between two **charged bodies** is **directly proportional** to the **product of electric charges** on the bodies and **inversely proportional** to the **square of the finite distance between the bodies**.

\[F=k\ \frac{q_1q_2}{r^2}\]

Where:

$F=$ **Electrostatic Force**

$q_1=$ **Charge of First Body**

$q_2=$ **Charge of Second Body**

$r=$ **Distance between two bodies**

$k=$ **Coulomb’s Constant** $=\ 9.0\times{10}^9\ \dfrac{N.m^2}{C^2}$

## Expert Answer

Given that:

**Distance between Sphere 1 and 2** $=r=20\ cm=20\times{10}^{-2}\ m$

**Electrostatic Force** $F=3.33\times{10}^{-21}\ N$

The** charge on both spheres is the same**, hence:

\[q_1=q_2=Q\]

First, we will find the **magnitude of electric charge** on both spheres by using **Coulomb’s Law**:

\[F\ =\ k\ \frac{q_1q_2}{r^2}\]

Since $q_1\ =\ q_2\ =\ Q$ , so:

\[F\ =\ k\ \frac{Q^2}{r^2}\]

By rearranging the equation:

\[Q=\ \sqrt{\frac{F\times r^2}{k}}\]

Substituting the given values in the above equation:

\[Q\ =\ \sqrt{\frac{(3.33\ \times\ {10}^{-21}\ N)\times{(20\ \times{10}^{-2}\ m)}^2}{\left(9.0\ \times\ {10}^9\ \dfrac{N.m^2}{C^2}\right)}}\]

\[Q\ =\ 1.22\ \times\ {10}^{-16}\ C\]

This is the **charge on both the spheres**.

Now, we will calculate the **excess electron** carried by spheres by using the formula for the **electric charge** as follows:

\[Q\ =\ n\times e\]

Where:

$Q\ =$ **Electric charge on the body**

$n\ =$ **Number of electrons**

$e\ =$ **Electric Charge on an electron** $=\ 1.602\ \times\ {10}^{-19}\ C$

So, by using the above formula:

\[n\ =\ \frac{Q}{e}\]

\[n\ =\ \frac{1.22\ \times\ {10}^{-16}\ C}{1.602\ \times\ {10}^{-19}\ C}\]

\[n\ =\ 0.7615\ \times\ {10}^3\]

\[n\ =\ 761.5\]

## Numerical Result

The **excess electrons** that each sphere carries to **repel** each other are $761.5$ **Electrons**.

## Example

Two bodies having an **equal and same charge** of $1.75\ \times\ {10}^{-16}\ C$ in space are **repelling** each other. If the bodies are separated by a **distance** of $60cm$, calculate the **magnitude of the repulsive force** acting between them.

**Solution**

Given that:

**Distance between two bodies** $=\ r\ =\ 60\ cm\ =\ 60\ \times{10}^{-2}\ m$

The** charge on both bodies is the same.** $q_1\ =\ q_2\ =\ 1.75\ \times\ {10}^{-16}\ C$

As per **Coulomb’s Law**, the **repulsive Electrostatic Force** is:

\[F\ =\ k\ \frac{q_1q_2}{r^2}\]

\[F\ =\ (9.0\ \times\ {10}^9\ \frac{N.m^2}{C^2})\ \frac{{(1.75\ \times\ {10}^{-16}\ C)}^2}{{(60\ \times{10}^{-2}\ m)}^2}\]

\[F\ =\ 7.656\times\ {10}^{-16}\ N\]