Question

The potential at a point $(x,0,0)$ is given by $V=(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3})$. The intensity of the electric field at $x=1$ will be

• A. $550V/m$
• B. $55V/m$
• C. $55000V/m$
• D. $5500V/m$

Answer 1

The correct option is D $5500V/m$

$V=(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3})$

We know that, $E=\frac{-dv}{dr}$

So,

$E=\frac{-d(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3})}{dx}$ at $x=1$

$=-[\frac{(-1000)}{x^2}-\frac{2\times 1500}{x^3}-\frac{3\times 500}{x^4}{]}_{x=1}$

$=1000+3000+1500=5500V/m$