Answer to Question #105551 in Electricity and Magnetism for happy456

Question

[img]https://upload.cc/i1/2020/03/13/9nKEaw.jpg[/img]




R=4

Answer 1

(a) The own energy of the shells

$W_1=\frac{q_1\phi_1}{2}=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_1^2}{2R_1}=$

$=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}\cdot\frac{(200\cdot10^{-12})^2}{2\cdot 0.15}\approx12\cdot10^{-10}J$

$W_2=\frac{q_2\phi_2}{2}=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_2^2}{2R_2}=$

$=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}\cdot\frac{(100\cdot10^{-12})^2}{2\cdot 0.24}\approx2\cdot10^{-10}J$

$W_3=\frac{q_3\phi_3}{2}=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_3^2}{2R_3}=$

$=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}\cdot\frac{(-450\cdot10^{-12})^2}{2\cdot 0.4}\approx23\cdot10^{-10}J$

The interaction energy of shells (the electric potential energy)

$W_i=W_{12}+W_{13}+W_{23}=\frac{1}{4\pi\epsilon_0}(q_1\frac{q_2+q_3}{a}+q_2\frac{q_1+q_3}{a}+q_3\frac{q_1+q_2}{a})=$

$=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}(200\cdot10^{-12}\frac{100\cdot10^{-12}-450\cdot10^{-12}}{3}+$

$+100\cdot10^{-12}\frac{200\cdot10^{-12}-450\cdot10^{-12}}{3}-450\cdot10^{-12}\frac{200\cdot10^{-12}+100\cdot10^{-12}}{3})=$

$=-8\cdot10^{-10}J$

The total energy of the shells

$W_{total}=W_1+W_2+W_3+W_{12}+W_{23}+W_{13}=29\cdot10^{-10}J$

(b)

$\phi_x=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_2}{R_2}=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}\cdot \frac{100\cdot10^{-12}}{0.24}=3.75 V$

$\phi_y=\frac{1}{4\pi\epsilon_0}\cdot\frac{q_3}{R_3}=\frac{1}{4\cdot 3.14\cdot 8.85\cdot10^{-12}}\cdot \frac{-450\cdot10^{-12}}{0.4}=-10.12 V$

$U_{xy}=\phi_x-\phi_y=3.75-(-10.12)=13.87V.$