Question:
The electric potential at any point (x, y, z) in metres is given by $V = 3x^{2}$. The electric field at a point (2, 0, 1) is
The electric potential at any point (x, y, z) in metres is given by $V = 3x^{2}$. The electric field at a point (2, 0, 1) is
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Electric field is the negative derivative of potential with respect to distance.
Updated On: Apr 10, 2026
- $12~Vm^{-1}$
- $-6~vm^{-1}$
- $6~Vm^{-1}$
- $-12~Vm^{-1}$
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The Correct Option is D
Solution and Explanation
Step 1: Relation
$E = -\frac{dV}{dx}$.
Step 2: Differentiation
Given $V = 3x^2$, then $E = -\frac{d}{dx}(3x^2) = -6x$.
Step 3: Substitution
At point (2, 0, 1), $x = 2$. Therefore, $E = -6(2) = -12~Vm^{-1}$.
Final Answer: (d)
$E = -\frac{dV}{dx}$.
Step 2: Differentiation
Given $V = 3x^2$, then $E = -\frac{d}{dx}(3x^2) = -6x$.
Step 3: Substitution
At point (2, 0, 1), $x = 2$. Therefore, $E = -6(2) = -12~Vm^{-1}$.
Final Answer: (d)
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