Question

An expanding sphere, radius$\mathit{R}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathit{v}\mathit{t}$(, constant) carries a charge Q, uniformly distributed over its volume. Evaluate the integral

${\mathit{Q}}_{\mathbf{e}\mathbf{f}\mathbf{f}}\mathbf{=}\int \rho (r,t)d\tau$

with respect to the center. Show that${\mathit{Q}}_{\mathbf{e}\mathbf{f}\mathbf{f}}\mathbf{\approx }\mathit{Q}\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{3}\mathbf{c}}{\mathbf{4}}\mathbf{)}$, if$\mathit{v}\mathbf{<}\mathbf{<}\mathit{c}$ .

Short answer

The expression for the required integral value for$v<<<c$ is $Q(1-\frac{3v}{4c})$and the value of the integral with respect to the centre is $Q_{eff}=\frac{3Q{c}^3}{v^3}[\ln \left(\frac{c+v}{c}\right)+\frac{v^2-2cv}{2c^2}]$.

Step-by-step solution

write the given data from the question.

The radius of the sphere,$R\left(t\right)=vt$

The charge of the sphere is .$Q$

Determine the formula to evaluate the line integral.

The expression for the volume charge density is given as follows.

$\mathit{\rho }\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$

Here,$\mathit{V}$ is the volume of the charge distribution.

The expression for the volumeof the sphere is given as follows.

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{R}}^{\mathbf{3}}$

Evaluate the line integral.

The volume charge density is given by,

$\rho (r,t)=\frac{Q}{V}$

Substitute$\frac{4}{3}\pi R^3$for $V$into above equation.

$\begin{array}{l}\rho (r,t)=\frac{Q}{\frac{4}{3}\pi R^3}\\ \rho (r,t)=\frac{3Q}{4\pi R^3}\end{array}$

Substitute$v{t}_r$for $R$into above equation.

$\rho (r,t)=\frac{3Q}{4\pi {\left(v{t}_r\right)}^3}$ …… (1)

For $r<R$the density of $r$is as follows.

$\begin{array}{l}{t}_r=t-\frac{r}{c}\\ t_r=\frac{tc-r}{c}\end{array}$

Substitute$\frac{tc-r}{c}$for$t_r$into equation (1).

$\begin{array}{l}\rho (r,t)=\frac{3Q}{4\pi {\left(v\left(\frac{tc-r}{c}\right)\right)}^3}\\ \rho (r,t)=\frac{3Q{c}^3}{4\pi v^3}\left(\frac{1}{{(tc-r)}^3}\right)\end{array}$

The given line integral,

$Q_{eff}={\int }_0^{vt}\rho (r,t)d\tau$

Substitute$\frac{3Q{c}^3}{4\pi v^3}\left(\frac{1}{{(tc-r)}^3}\right)$for$\rho (r,t)$into above equation.

$\begin{array}{l}{Q}_{eff}={\int }_0^{vt}\frac{3Q{c}^3}{4\pi v^3}\left(\frac{1}{{(tc-r)}^3}\right)d\tau \\ Q_{eff}=\frac{3Q{c}^3}{v^3}{\int }_0^{vt}\left(\frac{r^2}{{(tc-r)}^3}\right)dr\end{array}$

Here, the integration over the entire sphere and$\theta$,$\varphi$integral give the factor of $4\pi$.

$\begin{array}{l}{Q}_{eff}=\frac{3Q{c}^3}{v^3}{\int }_0^{vt}\left(\frac{r^2}{{(tc-r)}^3}\right)dr\\ Q_{eff}=\frac{3Q{c}^3}{v^3}[\ln \left(\frac{c}{c-v}\right)+\frac{3v^2-2cv}{2{(c-v)}^2}]\end{array}$

For ,$t>0$the expression can be reduced as,

$Q_{eff}=Q(1+\frac{9v}{4c})$

The above expression is not matched with the required result.

Similarly, for$R=vtV<<<c$

Now,

$\begin{array}{c}v{t}_r=c(t-t_r)\\ v{t}_r=ct-c{t}_r\\ t_r(c+v)=ct\\ t_r=\frac{ct}{(c+v)}\end{array}$

Calculate the expression for the radius as,

$R=v{t}_r$

Substitute $\frac{ct}{(c+v)}$for $t_r$into above equation.

$\begin{array}{l}v{t}_r=v\frac{ct}{(c+v)}\\ v{t}_r=\frac{vct}{(c+v)}\end{array}$

The integral is given by,

$Q_{eff}={\int }_0^{vt}\rho (r,t)d\tau$

Substitute$\frac{vct}{(c+v)}$for$v{t}_r$and $\frac{3Q{c}^3}{4\pi v^3}\left(\frac{1}{{(tc-r)}^3}\right)$for$\rho (r,t)$into above equation.

$Q_{eff}=\frac{3Q{c}^3}{v^3}{\int }_0^{\frac{vct}{c+v}}\left(\frac{r^2}{{(ct-r)}^3}\right)dr$

Here, the integration over the entire sphere and $\theta$, $\varphi$integral give the factor of $4\pi$.

$Q_{eff}=\frac{3Q{c}^3}{v^3}[\ln \left(\frac{c+v}{c}\right)+\frac{v^2-2cv}{2c^2}]$

Apply the condition $v<<<c$.

$Q_{eff}=(1-\frac{3v}{4c})$

Hence the expression for the required integral value for$v<<<c$ is$Q(1-\frac{3v}{4c})$ .