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 find the electric field at the point \((1\text{ m}, 0, 2\text{ m})\), we start with the given expression for the electric potential:

\(V = 4xz^2\) 

The electric field \((\mathbf{E})\) is related to the electric potential \((V)\) by the relation:

\(\mathbf{E} = -\nabla V\)

Here, the gradient \((\nabla V)\) in Cartesian coordinates is given by:

\(\nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right)\)

We calculate each of these partial derivatives:

Partial derivative with respect to \(x\):

\(\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(4xz^2) = 4z^2\)

Partial derivative with respect to \(y\):

\(\frac{\partial V}{\partial y} = 0\)

(since \(V\) does not depend on \(y\))

Partial derivative with respect to \(z\):

\(\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(4xz^2) = 8xz\)

Thus, the gradient of \(V\) becomes:

\(\nabla V = (4z^2, 0, 8xz)\)

Substituting these into the expression for the electric field, we have:

\(\mathbf{E} = -\nabla V = -(4z^2, 0, 8xz)\)

This means,

\(\mathbf{E} = (-4z^2, 0, -8xz)\)

We substitute \(x = 1\text{ m}\), \(y = 0\text{ m}\), \(z = 2\text{ m}\):

\(\mathbf{E} = (-4(2)^2, 0, -8(1)(2)) = (-16, 0, -16)\)

Thus, the electric field vector at \((1\text{ m}, 0, 2\text{ m})\) is \((-16, 0, -16)\, \text{V m}^{-1}\)

Evaluating the vector, we observe that the electric field has components only in the \(x\) and \(z\) directions, both negative. Thus, the direction of the electric field is along the line between the negative \(x\)-axis and negative \(z\)-axis.

From the options provided, the most accurate description of a single direction is:

Correct Answer: along the negative \(x\)-axis