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: \[\mathbf{E} = \bold{i}(2xy + z^3) + \bold{j}(x^2) + \bold{k}(3xz^2)\]

Key concepts

Gradient

The concept of a gradient is pivotal in vector calculus, as it generalizes the idea of a derivative to multi-variable functions. When we talk about the gradient of a function, we are essentially talking about a vector that points in the direction of the greatest rate of increase of that function. For a scalar function like our electric potential, the gradient is a vector field that shows how the potential changes in space.
The gradient of a function, such as the electric potential \[ V(x, y, z) = -x^2y - xz^3 + 4 \], is calculated by taking its partial derivatives with respect to each variable. The resulting vector is comprised of these partial derivatives:
\[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \].
This operation gives us the rate and direction of change of the potential for each of the independent variables \( x, y, \) and \( z \). The negative of this gradient gives us the electric field, which represents how a positive charge would experience a force in the potential field.

Partial Derivatives

Partial derivatives are at the heart of understanding how functions behave when they depend on multiple variables. Unlike ordinary derivatives, which deal with functions of a single variable, partial derivatives consider how a function changes as just one of those variables changes, while keeping the others constant.
For our electric potential function \( V(x, y, z) = -x^2y - xz^3 + 4 \), calculating the partial derivatives with respect to each variable allows us to understand how changes in \( x \), \( y \), and \( z \) independently affect the potential.

• The partial derivative with respect to \( x \) is \( -2xy - z^3 \).
• For \( y \), it is \( -x^2 \).
• And for \( z \), it is \( -3xz^2 \).

By finding these partial derivatives, we're essentially identifying the slope of the tangent line to the potential surface, viewed in the plane of each variable pair, one step closer to understanding the entire multidimensional landscape of the electric potential field.

Vector Calculus

Vector calculus provides the mathematical tools to handle functions and fields that have vectors as outputs and inputs. It's particularly important in physics for describing various vector fields, such as electric fields, magnetic fields, and force fields.
When dealing with electric fields derived from potential functions, vector calculus helps us move from the scalar potential to the vector field. This is done using operations like the gradient described earlier.
For example, given a potential function \( V(x, y, z) \), the electric field \( \mathbf{E} \) is obtained by taking the negative gradient of \( V \):\[ \mathbf{E} = -abla V \].
The result is a vector field representing the electric field. This means for every point in space, we have a vector pointing in the direction that a positive test charge would be accelerated towards.
The components of this field, found through vector calculus, not only give insight into the directionality but also magnitude of the electric forces, thus linking the abstract mathematical concepts directly to physical phenomena we can observe or measure.