Electric potential V at any point O ( x, y, z all in metre) in space is given by V=4x^2, volt . elec

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1. Electric Charges and Fields

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The electric potential $V$ at any point $O$ ($x, y, z$ all in metre) in space is given by $V=4x^2\, volt$. The electric field at the point $(1\,m, 0, 2\,m)$ in $volt/meter$ is

$8$ along negative $X -$ axis

$8$ along positive $X -$ axis

$16$ along negative $X -$ axis

$16$ along positive $Z -$ axis

Solution

The electric potential $\mathrm{V}(\mathrm{x}, \mathrm{y}, \mathrm{z})=4 \mathrm{x}^{2}$ $\mathrm{volt}$

Now $\overrightarrow {\rm{E}} = – \left( {\hat i\frac{{\partial {\rm{V}}}}{{\partial {\rm{x}}}} + \hat j\frac{{\partial {\rm{V}}}}{{\partial y}} + \widehat {\rm{k}}\frac{{\partial {\rm{V}}}}{{\partial {\rm{z}}}}} \right)$

Now $\frac{\partial \mathrm{V}}{\partial \mathrm{x}}=8 \mathrm{x}, \frac{\partial \mathrm{V}}{\partial \mathrm{y}}=0$ and $\frac{\partial \mathrm{V}}{\partial \mathrm{z}}=0$

Hence $\overrightarrow{\mathrm{E}}=-8 \mathrm{x} \hat{\mathrm{i}} .$ so at point $(1 \mathrm{\,m}, 0,2 \mathrm{\,m})$

$\overrightarrow {\rm{E}} = – 8\widehat {\rm{i}}$ $\mathrm{volt/metre}$ or $8$ along negative $\mathrm{X}$ – axis.

Standard 12

Physics

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Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. A positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

normal

The electric potential $V$ at any point $O$ ($x, y, z$ all in metre) in space is given by $V=4x^2\, volt$. The electric field at the point $(1\,m, 0, 2\,m)$ in $volt/meter$ is

Solution

Similar Questions

Electric field inside a uniformly charged sphere of radius $R,$ is ($r$ is distance from centre, $r < R$)

If the distance between two equal point charge is doubled then what would happen to the force between them ?

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