- Home
- Standard 12
- Physics
The electric potential $V$ at any point $O$ ($x$, $y$, $z$ all in metres) in space is given by $V = 4{x^2}\,volt$. The electric field at the point $(1m,\,0,\,2m)$ in $volt/metre$ is
$8$ along negative $X - $ axis
$8$ along positive $X - $ axis
$16$ along negative $X - $ axis
$16$ along positive $Z - $ axis
Solution
(a) The electric potential $V\,(x,y,z) = 4{x^2}\,volt$
Now $\overrightarrow E = – \,\left( {\hat i\frac{{\partial V}}{{\partial x}} + \hat j\frac{{\partial V}}{{\partial y}} + \hat k\frac{{\partial V}}{{\partial z}}} \right)$
Now $\frac{{\partial V}}{{\partial x}} = 8x,\,\frac{{\partial V}}{{\partial y}} = 0$ and $\frac{{\partial V}}{{\partial z}} = 0$
Hence $\overrightarrow E = – \,8x\hat i$, so at point ($1\,m, 0, 2\,m$)
$\overrightarrow E = – \,8\hat i\,\,volt/metre$ or $8$ along negative $X-$ axis.
