A hemispherical bowl of mass $m$ is uniformly charged with charge density $'\sigma '$ . Electric potential due to charge distribution at a point $'A'$ is (which lies at centre of radius as shown).

Define electric potential and explain it. Write its $\mathrm{SI}$ unit and give its other units.

The electric potential $V(x, y, z)$ for a planar charge distribution is given by: 

$V\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, < \, – d}\\
{ – {V_0}{{\left( {1 + \frac{x}{d}} \right)}^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\, – \,d\, \le x < 0}\\
{ – {V_0}\left( {1 + 2\frac{x}{d}} \right)\,\,\,\,\,\,\,\,\,\,\,for\,0\, \le x < d}\\
{ – 3{V_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, > \,d}
\end{array}} \right.$

where $-V_0$ is the potential at the origin and $d$ is a distance. Graph of electric field as a function of position is given as

The radius of a soap bubble whose potential is $16\,V$ is doubled. The new potential of the bubble will be…..$V$

Two equal positive point charges are kept at points $A$ and $B$ . The electric potential, while moving from $A$ to $B$ along straight line