Question: What is the radius of the imaginary concentric sphere that divides the electrostatic field of a meta...

Question

What is the radius of the imaginary concentric sphere that divides the electrostatic field of a metal sphere of a radius 20 cm & a charge of 8 mC in two regions of identical energy –

Answer 1

Solution

Energy between shell 1 & 2 = energy behind shell 2.

Energy between

shell 1 & 2 = $\int_{r = a}^{r = b}{\frac{1}{2} \in_{0}\left\lbrack \frac{KQ}{r^{2}} \right\rbrack^{2} \times 4\pi r^{2}dr}$

= $\frac{\in_{0}K^{2}Q^{2}}{2} \times 4\pi$ $\left\lbrack - \frac{1}{r} \right\rbrack_{a}^{b}$ = $\frac{\in_{0}K^{2}Q^{2}}{2}$ $\left\lbrack \frac{1}{a} - \frac{1}{b} \right\rbrack$

Energy beyond shell 2 = $\int_{r = b}^{r = \infty}{\frac{1}{2} \in_{0}}\left\lbrack \frac{KQ}{r^{2}} \right\rbrack^{2}$ × 4pr²dr

= $\frac{1}{2} \in_{0}K^{2}Q^{2}$ × 4p $\left\lbrack \frac{1}{b} \right\rbrack$

\ $\frac{1}{2} \in_{0}K^{2}Q^{2}$ × 4p $\left\lbrack \frac{1}{a} - \frac{1}{b} \right\rbrack$

= $\frac{1}{2} \in_{0}K^{2}Q^{2}$ × 4p $\left\lbrack \frac{1}{b} \right\rbrack$

$\frac{1}{a} = \frac{2}{b}$

b = 2a

a = 20 cm

b = 40 cm