Two charges $3 \times 10^{-8}\; C$ and $-2 \times 10^{-8}\; C$ are located $15 \;cm$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Two charges $3 \times 10^{-8}\; C$ and $-2 \times 10^{-8}\; C$ are located $15 \;cm$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Let us take the origin $O$ at the location of the positive charge. The line joining the two charges is taken to be the $x-$axis; the negative charge is taken to be on the right side of the origin
Let $P$ be the required point on the $x$ -axis where the potential is zero.
If $x$ is the $x$ -coordinate of $P$, obviously $x$ must be positive. (There is no possibility of potentials due to the two charges adding up to zero for $x < 0 .$ If $x$ lies between $O$ and $A$, we have
$\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{3 \times 10^{-8}}{x \times 10^{-2}}-\frac{2 \times 10^{-8}}{(15-x) \times 10^{-2}}\right]=0$
where $x$ is in $cm .$ That is,
$\frac{3}{x}-\frac{2}{15-x}=0$
which gives $x=9 cm$
If $x$ lies on the extended line $OA$, the required condition is $\frac{3}{x}-\frac{2}{x-15}=0$
which gives $x=45\; cm$
Thus, electric potential is zero at $9 cm$ and $45 cm$ away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.
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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 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