
- ✓$8.4$
- B$20$
- C$6.2$
- D$10.8$

$\Rightarrow \mathrm{E}=\frac{\frac{20}{4}+\frac{6}{3}}{\frac{1}{4}+\frac{1}{3}}=12$ $\mathrm{volt}$ $\Rightarrow \mathrm{r}=\frac{12}{7}$
$\Rightarrow \mathrm{V}=4\left[\frac{12}{4+\frac{12}{7}}\right]=8.4 \mathrm{\,volt}$
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$V\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, < \, - d}\\
{ - {V_0}{{\left( {1 + \frac{x}{d}} \right)}^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\, - \,d\, \le x < 0}\\
{ - {V_0}\left( {1 + 2\frac{x}{d}} \right)\,\,\,\,\,\,\,\,\,\,\,for\,0\, \le x < d}\\
{ - 3{V_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, > \,d}
\end{array}} \right.$
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
