The electric potential at any point (x, y, z) | vedclass

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Question

(D) The relation between electric field $\overrightarrow{E}$ and electric potential $V$ is given by $\overrightarrow{E} = -
abla V = -\left( \frac{\p

Vedclass · Accepted Answer

$-10\hat{i}$

Vedclass · Answer

(D) The relation between electric field $\overrightarrow{E}$ and electric potential $V$ is given by $\overrightarrow{E} = -
abla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} ight)$.Given $V = 5x^2$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = \frac{d}{dx}(5x^2) = 10x$$\frac{\partial V}{\partial y} = 0$$\frac{\partial V}{\partial z} = 0$Thus,$\overrightarrow{E} = -(10x)\hat{i} = -10x\hat{i} 	ext{ N/C}$.At the point $(1, 2, 3) 	ext{ m}$,substituting $x = 1$:$\overrightarrow{E} = -10(1)\hat{i} = -10\hat{i} 	ext{ N/C}$.Therefore,the correct option is $D$.

The electric potential at any point $(x, y, z)$ (all in meters) in space is given by $V = 5x^2$ volt. The electric field at the point $(1, 2, 3) \text{ m}$ is $\overrightarrow{E} = $ . . . . . . $\text{N/C}$.

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