Question

At any point $\left(\right.x,0,0\left.\right),$ the electric potential $V$ is $\left(\frac{10}{x} + \frac{15}{x^{2}} + \frac{5}{x^{3}}\right)$ volt, then electric field at $x=1m$ is

• A. $55\left(\right.\hat{j}+\hat{k}\left.\right)\lor /m$
• B. $55\hat{i}v/m$
• C. $\frac{55}{\sqrt{2}}\left(\right.\hat{i}+\hat{k}\left.\right)v/m$
• D. $55\hat{j}v/m$

Answer 1

Answer: (B)

$\overset{ \rightarrow }{E}=-\frac{dv}{dx}\hat{i}$
$=-\frac{dV}{dx}\left[\frac{10}{x} + \frac{15}{x^{2}} + \frac{5}{x^{3}}\right]\hat{i}$
$=-\left[- \frac{10}{x^{2}} - \frac{30}{x^{3}} - \frac{15}{x^{4}}\right]\hat{i}$
At $x=1$
$\overset{ \rightarrow }{E}=10+30+15$
$=55\hat{i}$