Question

The electric potential \(V\) at any point \((x,y,z)\) in space is given by \(V=4xz^2\) volt, where \(x,y,z\) are all in metre. The electric field at that point \((1\text{ m},0,2\text{ m})\) in \(\text{V m}^{-1}\) is

• A. along the positive \(z\)-axis
• B. along the positive \(x\)-axis
• C. along the positive \(y\)-axis
• D. along the negative \(z\)-axis
• E. along the negative \(x\)-axis

Hint:

Use: \[ \vec E=-\nabla V \] For a 3D potential, differentiate with respect to \(x\), \(y\), and \(z\) separately.

Answer 1

Answer: (E)

To determine the electric field at a point given the electric potential function, we use the concept that the electric field \(\vec{E}\) is the negative gradient of the electric potential \(V\). 

The electric potential at any point \((x, y, z)\) is provided as:

\(V = 4xz^2\)

The electric field can be found using the formula:

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

1. Calculate \(\frac{\partial V}{\partial x}\):
   • \(\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(4xz^2) = 4z^2\)
2. Calculate \(\frac{\partial V}{\partial y}\):
   • \(\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(4xz^2) = 0\)
3. Calculate \(\frac{\partial V}{\partial z}\):
   • \(\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(4xz^2) = 8xz\)

Substituting these partial derivatives, the electric field is:

\(\vec{E} = -(4z^2 \hat{i} + 0 \hat{j} + 8xz \hat{k})\)

Now, substitute the given point \((x, y, z) = (1, 0, 2)\) m:

• \(\vec{E} = -((4 \times 2^2) \hat{i} + 0 \hat{j} + (8 \times 1 \times 2) \hat{k}) = -(16 \hat{i} + 0 \hat{j} + 16 \hat{k})\)

Thus, at the point \((1\text{ m},0,2\text{ m})\), the electric field is \(-16 \hat{i} - 16 \hat{k}\) V/m.

Analyzing the direction, the components of the electric field show that it has a negative \(x\)- and \(z\)-component. Therefore, the electric field is along the negative \()-axis and negative \([z\)-axis.

Hence, the correct answer is "along the negative \(x\)-axis."