Question

The electric potential at points along x-direction is given by \(V = (1.0~\text{V/m}^2) x^2\). The electric field at point \(x = 5.0~\text{m}\) on x-axis is:

• A. -10.0 V/m
• B. +10.0 V/m
• C. -5.0 V/m
• D. +5.0 V/m

Hint:

Electric field is negative gradient of potential: \( \vec{E} = - \nabla V \). For 1D along x-axis, \(E_x = - dV/dx\).

Answer 1

The Correct Option is A

Step 1: Field–potential relationship.
E_x = -dV/dx.

Step 2: Differentiating the given function.
V = 1.0 x² → dV/dx = 2.0 x.

Step 3: Evaluating at x = 5 m.
E_x = -2.0 × 5 = -10 V/m.

Step 4: Conclusion.
The electric field at x = 5.0 m is -10.0 V/m.