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

• A. $12~Vm^{-1}$
• B. $-6~vm^{-1}$
• C. $6~Vm^{-1}$
• D. $-12~Vm^{-1}$

Hint:

Electric field is the negative derivative of potential with respect to distance.

Answer 1

The Correct Option is D

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)