Question

The electric potential V at any point (x, y, z), all in meters in space is given by $V=4x^2$ volt. The electric field at the point (1, 0, 2) in volt/meter, is :

• A. 8 along negative X-axis
• B. 8 along positive X-axis
• C. 16 along negative X-axis
• D. 16 along positive X-axis

Answer 1

The correct option is A 8 along negative X-axis

$\text{Step 1: Electric field at general point (x , y , z)}$

Given, Potential function as $V=4x^2$

Therefore, Electric field $E$ is given by:

$\overrightarrow{E}=\frac{-\partial V}{\partial x}\hat{i}-\frac{\partial V}{\partial y}\hat{j}-\frac{\partial V}{\partial z}\hat{k}$

$\overrightarrow{E}=\frac{-\partial }{\partial x}\left(4x^2\right)\hat{i}-\frac{\partial }{\partial y}\left(4x^2\right)\hat{j}-\frac{\partial }{\partial z}\left(4x^2\right)\hat{k}$

$\overrightarrow{E}=-8x\hat{i}-0-0$

$\overrightarrow{E}=-8x\hat{i}$

$\text{Step 2: Electric field at point (1 , 0 , 2) by putting value}$

$\overrightarrow{E}=-8x\hat{i}$ $=-8\left(1\right)\hat{i}$

$\overrightarrow{E}=-8\hat{i}$

Hence, electric field is $8N/C$ along negative x-axis. Therefore Option $A$