# What is the electric flux through a spherical surface just inside the inner surface of the sphere?

– 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 Gauss’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 static 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:

$E=\frac{\sigma}{\varepsilon_o}=\frac{Q}{A\times\varepsilon_o}$

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:

$\sigma_{new}=\sigma_1+\sigma_{out}$

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

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