Gujarati
Hindi
1. Electric Charges and Fields
normal

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

A

$8$ along negative $x-$ axis

B

$8$ along positive $x-$ axis

C

$16$ along negative $x-$ axis

D

$16$ along positive $x-$ axis

Solution

$\overrightarrow{\mathrm{E}}=-\left[\hat{\mathrm{i}} \frac{\partial \mathrm{V}}{\partial \mathrm{x}}+\hat{\mathrm{j}} \frac{\partial \mathrm{V}}{\partial \mathrm{y}}+\hat{\mathrm{k}} \frac{\partial \mathrm{V}}{\partial \mathrm{z}}\right]$

Given $\mathrm{V}=4 \mathrm{x}^{2}$

$\overrightarrow{\mathrm{E}}=-\hat{\mathrm{i}}(8 \mathrm{x})[$ Magnitude of $\overrightarrow{\mathrm{E}} \text { is }-8 \mathrm{\,V} / \mathrm{m}]$

Hence, electric field is $8 \mathrm{V} / \mathrm{m}$ along negative $\mathrm{X}$ -axis.

Standard 12
Physics

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