Chapter 22, Problem 85P

To determine

The electric field at point1 and point 2.

The electric field at point1 is $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point2 is $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Given:

A sphere of radius R and surface charge density $\rho$ .

A cavity inside the sphere of radius $b=R/2$ and charge Q.

Formula Used:

Gauss’s theorem

  ${\displaystyle {\oint }_S{E}_n}dA=\frac{Q_{inside}}{{\epsilon }_o}$

E is the electric field.

  ${\epsilon }_o$ is the permittivity of free space.

Q is the charge of the spere

Calculations:

The electric field by the sphere with cavity can be calculated as two uniform spheres of equal positive and negative charge densities.

Resultant electric field is

  $\overrightarrow{E}={\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }+{\overrightarrow{E}}_Q$

  ${\displaystyle {\oint }_S{E}_n}dA=\frac{Q_{inside}}{{\epsilon }_o}$

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $E_p(4\pi r^2)$

  $=\frac{Q_{inside}}{{\epsilon }_o}$

  $\rho =\frac{Q_{inside}}{\frac{4}{3}\pi r^3}$

  $Q_{inside}=\frac{4}{3}\pi r^3\rho$

Substituting the values

  $E_{\rho }=\frac{\rho r}{3{\epsilon }_o}$

  ${\overrightarrow{E}}_{\rho }=E_{\rho }\widehat{r}$

  ${\overrightarrow{E}}_{\rho }=\frac{\rho r}{3{\epsilon }_o}\widehat{r}$

Electric field due to negative charge density using gauss theorem

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $\frac{Q_{inside}}{{\epsilon }_o}$

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $E_{-\rho }(4\pi r{\text{'}}^2)$

  $=\frac{Q_{inside}}{{\epsilon }_o}$

Relating

  $-\rho =\frac{Q_{inside}}{\frac{4}{3}\pi r{\text{'}}^3}$

  $Q_{inside}=-\frac{4}{3}\pi r{\text{'}}^3\rho$

Substituting values

  $E_{-\rho }=-\frac{\rho r\text{'}}{3{\epsilon }_o}$

  $r\widehat{r}=\overrightarrow{r}=x\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $r\text{'}\widehat{r}\text{'}=\overrightarrow{r}\text{'}=(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Substituting in the equations

  ${\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }=\frac{\rho }{3{\epsilon }_o}(x\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})-\frac{\rho }{3{\epsilon }_o}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover> ]}$

  ${\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Calculating the electric field by the charge Q in the cavity

  ${\text{E}}_Q=\frac{Q}{4\pi {\epsilon }_o{b}^3}r\text{'}\widehat{r}\text{'}$

  $r\text{'}\widehat{r}\text{'}=\overrightarrow{r}\text{'}=(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_Q=\frac{Q}{4\pi {\epsilon }_o{b}^3}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

Resultant electric field $\overrightarrow{E}={\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }+{\overrightarrow{E}}_Q$

  $\overrightarrow{E}=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

At point 1

  ${\overrightarrow{E}}_1(2b,0)=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(2b-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

  $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

At point 2

  ${\overrightarrow{E}}_2(0,0)=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(0-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

  $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Conclusion:

The electric field at point1 is $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point2 is $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Answer to Problem 85P

The electric field at point1 is $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point2 is $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Explanation of Solution

Given:

A sphere of radius R and surface charge density $\rho$ .

A cavity inside the sphere of radius $b=R/2$ and charge Q.

Formula Used:

Gauss’s theorem

  ${\displaystyle {\oint }_S{E}_n}dA=\frac{Q_{inside}}{{\epsilon }_o}$

E is the electric field.

  ${\epsilon }_o$ is the permittivity of free space.

Q is the charge of the spere

Calculations:

The electric field by the sphere with cavity can be calculated as two uniform spheres of equal positive and negative charge densities.

Resultant electric field is

  $\overrightarrow{E}={\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }+{\overrightarrow{E}}_Q$

  ${\displaystyle {\oint }_S{E}_n}dA=\frac{Q_{inside}}{{\epsilon }_o}$

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $E_p(4\pi r^2)$

  $=\frac{Q_{inside}}{{\epsilon }_o}$

  $\rho =\frac{Q_{inside}}{\frac{4}{3}\pi r^3}$

  $Q_{inside}=\frac{4}{3}\pi r^3\rho$

Substituting the values

  $E_{\rho }=\frac{\rho r}{3{\epsilon }_o}$

  ${\overrightarrow{E}}_{\rho }=E_{\rho }\widehat{r}$

  ${\overrightarrow{E}}_{\rho }=\frac{\rho r}{3{\epsilon }_o}\widehat{r}$

Electric field due to negative charge density using gauss theorem

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $\frac{Q_{inside}}{{\epsilon }_o}$

  ${\displaystyle {\oint }_S{E}_n}dA=$

  $E_{-\rho }(4\pi r{\text{'}}^2)$

  $=\frac{Q_{inside}}{{\epsilon }_o}$

Relating

  $-\rho =\frac{Q_{inside}}{\frac{4}{3}\pi r{\text{'}}^3}$

  $Q_{inside}=-\frac{4}{3}\pi r{\text{'}}^3\rho$

Substituting values

  $E_{-\rho }=-\frac{\rho r\text{'}}{3{\epsilon }_o}$

  $r\widehat{r}=\overrightarrow{r}=x\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  $r\text{'}\widehat{r}\text{'}=\overrightarrow{r}\text{'}=(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

Substituting in the equations

  ${\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }=\frac{\rho }{3{\epsilon }_o}(x\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>})-\frac{\rho }{3{\epsilon }_o}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover> ]}$

  ${\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Calculating the electric field by the charge Q in the cavity

  ${\text{E}}_Q=\frac{Q}{4\pi {\epsilon }_o{b}^3}r\text{'}\widehat{r}\text{'}$

  $r\text{'}\widehat{r}\text{'}=\overrightarrow{r}\text{'}=(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}$

  ${\text{<mover accent="true"><mo>E</mo><mo>→</mo></mover>}}_Q=\frac{Q}{4\pi {\epsilon }_o{b}^3}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

Resultant electric field $\overrightarrow{E}={\overrightarrow{E}}_{\rho }+{\overrightarrow{E}}_{-\rho }+{\overrightarrow{E}}_Q$

  $\overrightarrow{E}=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(x-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+y\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

At point 1

  ${\overrightarrow{E}}_1(2b,0)=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(2b-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

  $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

At point 2

  ${\overrightarrow{E}}_2(0,0)=\frac{\rho b}{3{\epsilon }_o}\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover> +}\frac{Q}{4\pi {\epsilon }_o{b}^3}[(0-b)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}+0\text{<mover accent="true"><mi>j</mi> <mo>^</mo></mover>}]$

  $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

Conclusion:

The electric field at point1 is $E_1=\left(\frac{\rho b}{3{\epsilon }_o}+\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$

The electric field at point2 is $E_2=\left(\frac{\rho b}{3{\epsilon }_o}-\frac{Q}{4\pi {\epsilon }_o{b}^2}\right)\text{<mover accent="true"> <mi>i</mi> <mo>^</mo></mover>}$