Question

Consider a cube of uniform charge density ρ. The ratio of electrostatic potential at the centre of the cube to that at one of the corners of the cube is

Answer 1

$\begin{document}</p><p></p><p>\ch{</p><p> "\OX{o1,Zn}" \sld{} + "\OX{r1,Cu}" {}^2+ \aq</p><p> -></p><p> "\OX{r2,Cu}" \sld{} + "\OX{o2,Zn}" {}^2+ \aq</p><p>}</p><p>\redox(o1,o2)[->]{\small oxidation (2 electrons lost)}</p><p>\redox(r1,r2)[->][-1]{\small reduction (2 electrons gained)}</p><p></p><p>$

Answer 2

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ĄNsWeR࿐}}}$

$\begin{document}</p><p></p><p>\ch{</p><p> "\OX{o1,Zn}" \sld{} + "\OX{r1,Cu}" {}^2+ \aq</p><p> -></p><p> "\OX{r2,Cu}" \sld{} + "\OX{o2,Zn}" {}^2+ \aq</p><p>}</p><p>\redox(o1,o2)[->]{\small oxidation (2 electrons lost)}</p><p>\redox(r1,r2)[->][-1]{\small reduction (2 electrons gained)}</p><p></p><p>$

HOPE SO IT WILL HELP......