91影视

The differential energy due to differential charge,

$dU=V\left(r\right)dq$

A point charge is a sphere having a total charge (Q) and radius (R)

Potential due to a sphere, hence, the voltage is given as

$V\left(r\right)=\frac{kq}{r}$

Here the charge (q) is dependent on (r) as the charge is distributed uniformly over the sphere, so, the charge is given as

role="math" localid="1665122135288" $$\begin{gathered} q=蚁V_{ol}=\frac{4蚁{\mathrm{蟺谤}}^3}{3} \\ V\left(r\right)=\frac{4k蚁{\mathrm{蟺谤}}^3}{3r}=\frac{4k蚁{\mathrm{蟺谤}}^3}{3} \end{gathered}$$

As the charge element is changed with the volume,

Hence, the charge is given as the function in the volume dimension

$dq=蚁d{V}_{ol}=蚁\left(4{\mathrm{蟺谤}}^2\right)dr$

By substitution;

$$\begin{gathered} U=鈭�V\left(r\right)dq \\ ={鈭�}_0^R\left(\frac{4k{\mathrm{蟺辫谤}}^2}{3}\right)\left(4{\mathrm{蟺辫谤}}^2\right)dr \\ =\left(\frac{4^{2}k{蟺}^2{p}^2}{3}\right){\left[\frac{r^5}{5}\right]}_0^R \\ U=\left(\frac{4^{2}k{蟺}^2{p}^2}{3}\right)\left[\frac{R^5}{5}\right] \\ =3\left(\frac{4^{2}k{蟺}^2{p}^2}{3}\right)\left[\frac{R^6}{5R}\right] \\ =\left(\frac{3k}{5R}\right)\left(\frac{4蚁蟺R^3}{3}\right)\left(\frac{4蚁蟺R^3}{3}\right) \\ =\left(\frac{3k}{5R}\right)\left(蚁V\right)\left(蚁V\right) \\ U=\frac{3k{Q}^2}{5R} \end{gathered}$$

Putting the values,

$U=\frac{3k{Q}^2}{5R}=\left(\frac{3}{5}\right)\left(\frac{Q^2}{4{\mathrm{蟺蔚}}_0\mathrm{R}}\right)$

Hence, the energy added is Hence, the energy added is$U=\frac{3k{Q}^2}{5R}=\left(\frac{3}{5}\right)\left(\frac{Q^2}{4{\mathrm{蟺蔚}}_0\mathrm{R}}\right)$..