The electric potential V at any point (x, y, z), all in meters in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/ meter is ?

The electric potential V at any point (x, y, z), all in meters in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/ meter is ?

(a) 8 along positive X-axis

(b) 16 along negative X-axis

(c) 16 along positive X-axis

(d) 8 along negative X-axis

Answer :  (d) 8 along negative X-axis

\( V(x, y, z) = 4x^2 \) volts

Electric Field:

\[ \vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]

Compute Partial Derivatives:

\[ \frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x \] \[ \frac{\partial V}{\partial y} = 0 \] \[ \frac{\partial V}{\partial z} = 0 \]

Substitute into Electric Field Expression:

\[ \vec{E} = -\left(8x \hat{i} + 0 \hat{j} + 0 \hat{k} \right) = -8x \hat{i} \]

At the point (1, 0, 2):

\[ \vec{E} = -8(1)\hat{i} = -8 \hat{i} \text{ V/m} \]

Conclusion:

• Magnitude: \(8\)
• Direction: along negative X-axis