Question

Given the potential function in free space to be $V\left(x\right)=(50x^2+50y^2+50z^2)$ volts, the magnitude (in volts/metre) and the direction of the electric field at a point (1, -1, 1), where the dimensions are in meters, are

• A. $100;(\hat{i}+\hat{j}+\hat{k})$
  
• B. $\frac{100}{\sqrt{3}};(\hat{i}-\hat{j}+\hat{k})$
  
• C. $100\sqrt{3}$; $(\hat{i}+\hat{j}-\hat{k})/\sqrt{3}$
  
• D. $100\sqrt{3}$; $(-\hat{i}+\hat{j}-\hat{k})/\sqrt{3}$
  

Answer 1

The correct option is D $100\sqrt{3}$; $(-\hat{i}+\hat{j}-\hat{k})/\sqrt{3}$


Ans : (d)

$V_{x,y,z}=50x^2+50y^2+50z^2$
$\overrightarrow{E}=-▽V$
$=-[100x\hat{i}+100y\hat{j}+100z\hat{k}]V/m$
$\overrightarrow{E}(1,-1,1)=-[100\hat{i}-100\hat{j}+100\hat{k}]V/m$
$|E|=\sqrt{{100}^2+{100}^2+{100}^2}=100\sqrt{3}$
${\hat{a}}_E=\frac{\overrightarrow{E}}{|E|}=-\frac{1}{\sqrt{3}}(i-j+k)$