In space,the electric potential varies as V = | vedclass

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(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$.Given $V = 20|\vec{r}| = 20\sqrt{x^2 +

Vedclass · Accepted Answer

$-\sqrt{2}(8 \hat{i} + 6 \hat{j} - 10 \hat{k})$

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(B) The electric field $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -
abla V$.Given $V = 20|\vec{r}| = 20\sqrt{x^2 + y^2 + z^2}$.Calculating the gradient $
abla V = \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k}$.$\frac{\partial V}{\partial x} = 20 \cdot \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2x = \frac{20x}{|\vec{r}|}$.Similarly,$\frac{\partial V}{\partial y} = \frac{20y}{|\vec{r}|}$ and $\frac{\partial V}{\partial z} = \frac{20z}{|\vec{r}|}$.Thus,$\vec{E} = -\frac{20}{|\vec{r}|} (x \hat{i} + y \hat{j} + z \hat{k}) = -\frac{20}{|\vec{r}|} \vec{r} = -20 \hat{r}$.At the point $(4, 3, -5)$,the magnitude $|\vec{r}| = \sqrt{4^2 + 3^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.Substituting the values: $\vec{E} = -\frac{20}{5\sqrt{2}} (4 \hat{i} + 3 \hat{j} - 5 \hat{k}) = -\frac{4}{\sqrt{2}} (4 \hat{i} + 3 \hat{j} - 5 \hat{k}) = -2\sqrt{2} (4 \hat{i} + 3 \hat{j} - 5 \hat{k})$.This simplifies to $-\sqrt{2} (8 \hat{i} + 6 \hat{j} - 10 \hat{k})$.Therefore,the correct option is $B$.

In space,the electric potential varies as $V = 20|\vec{r}|$ volt,where $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ is the position vector. Then,the electric field in $N C^{-1}$ at the point $(4 \ m, 3 \ m, -5 \ m)$ is:

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