Question

The electric potential in a volume of space is given by \(V(x, y, z)=x^{2}+x y^{2}+y z\). Determine the electric field in this region at the coordinate (3,4,5) .

Short answer

The electric field at the point (3,4,5) is \(\vec{E}(3,4,5) = -22 \hat{i} - 29 \hat{j} - 4 \hat{k}\).

Step-by-step solution

Find the gradient of the electric potential function

To find the electric field, we first need to determine the gradient of the electric potential function. The gradient, denoted by \(\nabla V\), is a vector containing the partial derivatives of \(V(x, y, z)\) with respect to each spatial coordinate. In other words, we need to compute \(\frac{\partial V}{\partial x}\), \(\frac{\partial V}{\partial y}\), and \(\frac{\partial V}{\partial z}\).

Calculate the partial derivatives

Let's calculate the partial derivatives of the given electric potential function \(V(x, y, z) = x^2 + xy^2 + yz\). Partial derivative with respect to \(x\): \(\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} (x^2 + xy^2 + yz) = 2x + y^2\) Partial derivative with respect to \(y\): \(\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} (x^2 + xy^2 + yz) = 2xy + z\) Partial derivative with respect to \(z\): \(\frac{\partial V}{\partial z} = \frac{\partial}{\partial z} (x^2 + xy^2 + yz) = y\)

Write the gradient vector

Now that we have calculated all three partial derivatives, we can write the gradient vector, which represents the electric field \(\vec{E}\) in terms of \(x\), \(y\), and \(z\). \(\vec{E}(x, y, z) = -\nabla V(x, y, z) = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\right)\) Substitute the partial derivatives we calculated earlier to get: \(\vec{E}(x, y, z) = -(2x + y^2) \hat{i} - (2xy + z) \hat{j} - y \hat{k}\)

Determine the electric field at the given point (3,4,5)

We have the expression for the electric field \(\vec{E}(x, y, z)\), so now we can substitute the given point (3,4,5) into the electric field's coordinates to determine the electric field at that point: \(\vec{E}(3, 4, 5) = -(2(3) + (4)^2) \hat{i} - (2(3)(4) + 5) \hat{j} - 4 \hat{k}\) Calculating gives us: \(\vec{E}(3, 4, 5) = -(6 + 16) \hat{i} - (24 + 5) \hat{j} - 4 \hat{k} = -22 \hat{i} - 29 \hat{j} - 4 \hat{k}\) So, the electric field at the coordinate (3,4,5) is \(\vec{E}(3,4,5) = -22 \hat{i} - 29 \hat{j} - 4 \hat{k}\).

Key concepts

Electric Potential

Electric potential is a fundamental concept in electromagnetism that represents the potential energy per unit charge at a point in space. It's caused by electric charge and is measured in volts. Imagine you are pushing a small, positively-charged particle against an electric field; the work needed to move this particle from one point to another defines the electric potential difference between those points.

Using the formula given in our exercise, V(x, y, z) = x^2 + xy^2 + yz, we see a three-dimensional representation of electric potential, where each term incorporates the interactions between charges along the x, y, and z axes. When we talk about determining the electric field, we're essentially looking for the force that a charge would experience due to this potential.

Gradient of Electric Potential

The gradient of electric potential is a vector that points in the direction of the greatest rate of increase of the potential. It has both magnitude and direction, and in the context of electric fields, it's crucially important because it defines the electric field itself, with the electric field (E) being the negative gradient of the electric potential (V).

In vector calculus, the gradient transforms a scalar field, like electric potential, into a vector field. In our example, finding the gradient of V(x, y, z) helps us determine the electric field at any point in that space. The negative sign in the electric field definition comes from the fact that electric fields do work against the electric potential.

Partial Derivatives

Partial derivatives play a central role in understanding how a function changes as each variable is varied independently. In our problem, we're analyzing an electric potential that depends on x, y, and z; thus, we find partial derivatives with respect to each coordinate.

The notation ∂V/∂x signifies how much V changes as x changes while keeping y and z constant. Similarly, ∂V/∂y and ∂V/∂z describe changes along the y and z axes, respectively. These derivatives reveal how a specific component of the electric potential changes in a specific direction, giving us a granular view of its behavior in space.

Vector Calculus

Vector calculus is a branch of mathematics dedicated to the study of vectors in multiple dimensions. It extends single-variable calculus (dealing with functions of one variable) into higher dimensions. This field includes operations like gradient, divergence, and curl.

The process of determining the gradient of the electric potential function and subsequently the electric field involves applying vector calculus principles. This methodology adds depth to our understanding by accounting for multi-directional changes and interrelationships between spatial components. Thus, while the electric potential function describes the energy landscape, vector calculus provides us with the tools to navigate and describe the force experienced by charges within this landscape.