**– A conducting sphere with a hollow cavity inside has an outer radius of $0.250m$ and an internal radius of $0.200m$. A uniform charge exists on its surface having a density of $+6.37\times{10}^{-6}\dfrac{C}{m^2}$. Inside the cavity of the sphere, a new charge having a magnitude of $-0.500\mu C$ is introduced.**

**– (a) Calculate the new charge density that is developed on the outer surface of the sphere.**

**– (b) Calculate the electric field strength that exists on the outside of the sphere.**

**– (c) On the inside surface of the sphere, calculate the electric flux that is passing through the spherical surface.**

The aim of this article is to find the **surface charge density** $\sigma$, **electric field** $E$, and **electric flux** $\Phi$ induced by **electric charge** $Q$.

The basic concept behind this article is **Gaussâ€™s Law for Electric field**, **Surface Charge Density** $\sigma$, and **Electrical Flux** $\Phi$.

**Gauss’s law for the electric field** is the representation of the s**tatic electric field** which is created when **electrical charge** $Q$ is distributed across the **conducting surface** and the **total electrical flux** $\Phi$ passing through a **charged surface** is expressed as follows:

\[\Phi=\frac{Q}{\varepsilon_o}\]

**Surface Charge Density** $\sigma$ is the distribution of **electrical charge** $Q$ **per unit area** $A$ and is represented as follows:

\[\sigma=\frac{Q}{A}\]

The **strength of Electrical Field** $E$ is expressed as:

\[E=\frac{\sigma}{\varepsilon_o}=\frac{Q}{A\times\varepsilon_o}\]

## Expert Answer

Given that:

**Internal Radius of the sphere** $r_{in}=0.2m$

**Outer Radius of the sphere** $r_{out}=0.25m$

**Initial Surface Charge Density** on sphere surface $\sigma_1=+6.37\times{10}^{-6}\dfrac{C}{m^2}$

**Charge inside the cavity** $Q=-0.500\mu C=-0.5\times{10}^{-6}C$

**Area of the sphere** $A=4\pi r^2$

**Permittivity of Free Space** $\varepsilon_o=8.854\times{10}^{-12}\dfrac{C^2m^2}{N}$

**Part (a)**

**Charge Density** on the **outer surface** of the **sphere** is:

\[\sigma_{out}=\frac{Q}{A}=\frac{Q}{4\pi{r_{out}}^2}\]

\[\sigma_{out}=\frac{-0.5\times{10}^{-6}C}{4\pi{(0.25m)}^2}\]

\[\sigma_{out}=-6.369\times{10}^{-7}\frac{C}{m^2}\]

The **Net Charge Density** $\sigma_{new}$ on the **outer surface** after **charge** introduction is:

\[\sigma_{new}=\sigma_1+\sigma_{out}\]

\[\sigma_{new}=6.37\times{10}^{-6}\frac{C}{m^2}+(-6.369\times{10}^{-7}\frac{C}{m^2})\]

\[\sigma_{new}=5.733\times{10}^{-6}\frac{C}{m^2}\]

**Part (b)**

The **strength of Electrical Field** $E$ is expressed as:

\[E=\frac{\sigma}{\varepsilon_o}\]

\[E=\frac{5.733\times{10}^{-6}\dfrac{C}{m^2}}{8.854\times{10}^{-12}\dfrac{C^2m^2}{N}}\]

\[E=6.475\times{10}^5\frac{N}{C}\]

**Part (c)**

The **electric flux** $\Phi$ that is passing through the **spherical surface** after the introduction of** charge** $Q$ is expressed as:

\[\Phi=\frac{Q}{\varepsilon_o}\]

\[\Phi=\frac{-0.5\times{10}^{-6}C\ }{8.854\times{10}^{-12}\dfrac{C^2m^2}{N}}\]

\[\Phi=-5.647{\times10}^4\frac{Nm^2}{C}\]

## Numerical Result

**Part (a)** – The **Net Surface Charge Density** $\sigma_{new}$ on the **outer surface** of the **sphere** after **charge** introduction is:

\[\sigma_{new}=5.733\times{10}^{-6}\frac{C}{m^2}\]

**Part (b)** – The **strength of Electrical Field** $E$ that exists on the **outside** of the **sphere** is:

\[E=6.475\times{10}^5\frac{N}{C}\]

**Part (c)** – The **electric flux** $\Phi$ that is passing through the **spherical surface** after the introduction of **charge** $Q$ is:

\[\Phi=-5.647{\times10}^4\frac{Nm^2}{C}\]

## Example

A **conducting sphere** with a **cavity** inside has an **outer radius** of $0.35m$. A **uniform charge** exists on its **surface** having a **density** of $+6.37\times{10}^{-6}\frac{C}{m^2}$. Inside the cavity of the sphere, a **new charge** having a magnitude of $-0.34\mu C$ is introduced. Calculate the **new** **charge density** that is developed on the **outer surface** of the **sphere**.

**Solution**

Given that:

**Outer Radius **$r_{out}=0.35m$

**Initial Surface Charge Density** **on sphere surface** $\sigma_1=+6.37\times{10}^{-6}\dfrac{C}{m^2}$

**Charge inside the cavity** $Q=-0.34\mu C=-0.5\times{10}^{-6}C$

**Area of the sphere** $A=4\pi r^2$

**Charge Density** on the **outer surface** of the **sphere** is:

\[\sigma_{out}=\frac{Q}{A}=\frac{Q}{4\pi{r_{out}}^2}\]

\[\sigma_{out}=\frac{-0.34\times{10}^{-6}C}{4\pi{(0.35m)}^2}\]

\[\sigma_{out}=-2.209\times{10}^{-7}\frac{C}{m^2}\]

The **Net Charge Density** $\sigma_{new}$ on the** outer surface** after **charge** introduction is:

\[\sigma_{new}=\sigma_1+\sigma_{out}\]

\[\sigma_{new}=6.37\times{10}^{-6}\frac{C}{m^2}+(-2.209\times{10}^{-7}\frac{C}{m^2})\]

\[\sigma_{new}=6.149\times{10}^{-6}\frac{C}{m^2}\]