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2. Electric Potential and Capacitance
hard
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
A
B
C
D
Solution
$\mathrm{E}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}}$
Standard 12
Physics
