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  4. The electric potential V is given as a function of distance x (metre) by V=(5 x2+10 x-9) volt. Value of electric field at x=1 is

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Q. The electric potential $V$ is given as a function of distance $x$ (metre) by $V=\left(5 x^{2}+10 x-9\right)$ volt. Value of electric field at $x=1$ is

Electrostatic Potential and Capacitance

Solution:

$E=-\frac{d V}{d x}$
$=-\frac{d}{d x}\left(5 x^{2}+10 x-9\right)$
$=-10 x-10$
$\therefore (E)_{x=1}=-10 \times 1-10$
$=-20\, V / m$