Question

At any point $\left(x , \, 0 , \, 0\right)$ the electric potential $V$ is $\left(\frac{1000}{x} + \frac{1500}{x^{2}} + \frac{500}{x^{3}}\right) \, volt$ , then electric field intensity at $x=1 \, m$ is

• A. $5500\left(\hat{j} + \hat{k}\right)Vm^{- 1}$
• B. $5500\hat{i}Vm^{- 1}$
• C. $\frac{5500}{\sqrt{2}}\left(\hat{j} + \hat{k}\right)Vm^{- 1}$
• D. $\frac{5500}{\sqrt{2}}\left(\hat{i} + \hat{k}\right)Vm^{- 1}$

Answer 1

Answer: (B)

Electric potential
$V=\left(\frac{1000}{x} + \frac{1500}{x^{2}} + \frac{500}{x^{3}}\right)$
Electric field
$E=-\frac{dV}{dx}=\left(\frac{1000}{x^{2}} + \frac{\left(2\right) 1500}{x^{3}} + \frac{\left(3\right) 500}{x^{4}}\right)\hat{i}$
At $x=1$ ,
$E=5500\hat{i} \, Vm^{- 1}$