Question

The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)

Short answer

The electric field at point P, \(\mathbf{E}(x, y, z) = \left(2xy + z^3, x^2, 3xz^2\right)\), corresponds to the answer (A) and can be represented as: \[\mathbf{E} = \bold{i}(2xy + z^3) + \bold{j}(x^2) + \bold{k}(3xz^2)\]

Step-by-step solution

Identify the electric potential function

Here, the electric potential function \(V(x, y, z)\) is given by: \[V(x, y, z) = -x^2y - xz^3 + 4\]

Calculate the gradient of V

To find the electric field, we need to compute the gradient of V, i.e., its derivative along \(x\), \(y\), and \(z\) directions. The gradient of V is given as: \(\nabla V =\left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z}\right)\). Now, let's compute each partial derivative: \(\frac{\partial V}{\partial x} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial x} = -2xy - z^3\) \(\frac{\partial V}{\partial y} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial y} = -x^2\) \(\frac{\partial V}{\partial z} = \frac{\partial(-x^2y - xz^3 + 4)}{\partial z} = -3xz^2\) So, the gradient of V is: \(\nabla V =\left(-2xy - z^3, -x^2, -3xz^2\right)\)

Determine the electric field using the gradient of V

We know that the electric field \(\mathbf{E}\) is given by: \[\mathbf{E} = -\nabla V\] Substituting the values of the gradient from step 2, we get: \[\mathbf{E} = -(-2xy - z^3, -x^2, -3xz^2) = \left(2xy + z^3, x^2, 3xz^2\right)\] So, the electric field at point P, \(\mathbf{E}(x, y, z) = \left(2xy + z^3, x^2, 3xz^2\right)\). This corresponds to the answer (A), which is: \[\mathbf{E} = \bold{i}(2xy + z^3) + \bold{j}(x^2) + \bold{k}(3xz^2)\]