- Home
- Standard 12
- Physics
The potential at a point $x$ (measured in $μ\ m$) due to some charges situated on the $ x$-axis is given by $V(x)$ =$\frac{{20}}{{{x^2} - 4}}$ $volt$ The electric field $E$ at $x = 4\ μ m$ is given by
$\frac{{10}}{9}$ $\frac{V}{{\mu m}}$ and in the $+ve\ x$ direction
$\;\frac{5}{3}$ $\frac{V}{{\mu m}}$ and in the $-ve\ x$ direction
$\;\frac{5}{3}$ $\frac{V}{{\mu m}}$ and in the $+ve\ x$ direction
$\;\frac{{10}}{9}$ $\frac{V}{{\mu m}}$ and in the $-ve\ x$ direction
Solution
Here, $V(x)=\frac{20}{x^{2}-4}$ $volt$
We know that $E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^{2}-4}\right)$
or, $E=+\frac{40 x}{\left(x^{2}-4\right)^{2}}$
At $x=4\, \mu \mathrm{m}$
$E=+\frac{40 \times 4}{\left(4^{2}-4\right)^{2}}=+\frac{160}{144}=+\frac{10}{9}$ $volt / \mu \mathrm{m}$
Positive sign indicates that $\vec{E}$ is in $+ ve\, x-$ direction.
