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

• A. $-\sqrt{2} (4i + 3j - 10k)$
• B. $-\sqrt{2} (8i + 6j - 10k)$
• C. $-(8i + 6j - 10k)$
• D. $4i + 3j - 5k$

Hint:

Use the gradient operator carefully when dealing with electric fields and potentials in vector form.

Answer 1

The Correct Option is B

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