Question

The electric potential \( V(x) \) in a region around the origin is given by \( V(x) = 4x^2 \, {volts}. \) .The electric charge enclosed in a cube of 1m side with its center at the origin is (in coulomb)

• A. \( 8\epsilon_0 \)
• B. \( -4\epsilon_0 \)
• C. 0
• D. \( -8\epsilon_0 \)

Hint:

The electric field can be derived from the electric potential by taking the negative gradient. For a simple case like \( V(x) = 4x^2 \), the electric field is linear and can be integrated to find the enclosed charge.

Answer 1

The Correct Option is D

Step 1: The electric field \( E \) is related to the potential \( V \) by the relation: \[ E = -\frac{dV}{dx} \] Given \( V(x) = 4x^2 \), the electric field is: \[ E = -\frac{d}{dx}(4x^2) = -8x. \] Step 2: The charge enclosed by a surface can be calculated using Gauss's Law: \[ Q_{{enc}} = \oint E \, dA = \int E \, dA. \] Since the electric field varies with \( x \), we integrate over the volume of the cube. 
Step 3: After performing the integration, the charge enclosed in the cube is: \[ Q_{{enc}} = -8\epsilon_0. \]