Question:
In space, the electric potential varies as $V = 20|r|$ volt, where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector. Then electric field in (N/C) at the point (4 m, 3 m, -5 m) is
In space, the electric potential varies as $V = 20|r|$ volt, where $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector. Then electric field in (N/C) at the point (4 m, 3 m, -5 m) is
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Use the gradient operator carefully when dealing with electric fields and potentials in vector form.
Updated On: Mar 16, 2026
- $-\sqrt{2} (4i + 3j - 10k)$
- $-\sqrt{2} (8i + 6j - 10k)$
- $-(8i + 6j - 10k)$
- $4i + 3j - 5k$
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The Correct Option is B
Solution and Explanation
The electric field is the negative gradient of the electric potential:
$$ \vec{E} = - \nabla V $$
Given that $V = 20|r|$, the gradient of $V$ is $\nabla V = 20 \hat{r}$, where $\hat{r}$ is the unit vector in the direction of $\vec{r}$. Substituting the coordinates $(4, 3, -5)$, we get the electric field as $-\sqrt{2} (8i + 6j - 10k)$.
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