Answer to Question #106257 in Electricity and Magnetism for havefun7741

Question

[img]https://upload.cc/i1/2020/03/23/iuJgmO.jpg[/img]



the question is in picture,R=4

Answer 1

$\phi=\phi_1+\phi_2+\phi_3+\phi_4$

For two point charges $\phi_1=\phi_2=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r}=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{(d/\cos30°)}=$

$=\frac{1}{4\cdot3.14\cdot8.85\cdot10^{-12}}\cdot\frac{2\cdot10^{-9}}{(0.1/\cos30°)}\approx156$ $V$

For the linear rod

$\phi_3=\frac{\lambda}{4\pi\epsilon_0}\cdot \ln{\frac{b+\sqrt{4a^2+b^2} }{-b+\sqrt{4a^2+b^2}}}=\frac{-10\cdot10^{-9}}{4\cdot3.14\cdot8.85\cdot10^{-12}}\cdot \ln{\frac{0.08+\sqrt{4\cdot0.05^2+0.08^2} }{-0.08+\sqrt{4\cdot 0.05^2+0.08^2}}}\approx-132$ $V$

For the thin circular rod

$\phi_4=\frac{1}{4\pi\epsilon_0}\cdot\frac{2\pi\cdot R\cdot \lambda}{R}=\frac{\lambda}{2\epsilon_0}=\frac{5\cdot10^{-9}}{2\cdot8.85\cdot10^{-12}}\approx282$ $V$

$\phi=\phi_1+\phi_2+\phi_3+\phi_4=156+156-132+282=462$ $V$