Electric Field Intensity Calculation From Potential V = -2xy

In the realm of electromagnetism, the concepts of electric potential and electric field intensity are fundamental to understanding the behavior of charged particles and their interactions. Electric potential, often denoted by V, represents the amount of work required to move a unit positive charge from a reference point to a specific location in an electric field. It is a scalar quantity measured in volts (V). Electric field intensity, on the other hand, denoted by E, is a vector quantity that describes the force exerted on a unit positive charge at a given point in space. It is measured in Newtons per Coulomb (N/C).

The relationship between electric potential and electric field intensity is crucial. The electric field intensity is the negative gradient of the electric potential. This means that the electric field points in the direction of the steepest decrease in electric potential. Mathematically, this relationship is expressed as:

E = -∇V

Where ∇ is the gradient operator. In Cartesian coordinates, the gradient operator is given by:

∇ = (∂/∂x)i + (∂/∂y)j + (∂/∂z)k

Where i, j, and k are the unit vectors in the x, y, and z directions, respectively.

This relationship allows us to determine the electric field intensity if we know the electric potential as a function of position. Conversely, we can determine the electric potential if we know the electric field intensity.

In this article, we will explore a specific example where the electric potential at a point P(x, y) in the x-y plane is given by V = -2xy. Our goal is to determine the electric field intensity at the point (1, -1). This example will provide a practical application of the relationship between electric potential and electric field intensity.

By understanding these concepts and their relationship, we can gain a deeper understanding of the behavior of electric fields and their interactions with charged particles. This knowledge is essential in various fields, including electronics, telecommunications, and medical imaging.

Let's delve into the problem at hand. We are given the electric potential V at a point P(x, y) in the x-y plane as:

V = -2xy

Our objective is to determine the electric field intensity E at the specific point (1, -1). To achieve this, we will utilize the relationship between electric potential and electric field intensity, which states that the electric field intensity is the negative gradient of the electric potential:

E = -∇V

Where ∇ is the gradient operator. In two dimensions (x-y plane), the gradient operator is given by:

∇ = (∂/∂x)i + (∂/∂y)j

Where i and j are the unit vectors in the x and y directions, respectively.

To find the electric field intensity, we need to calculate the partial derivatives of the electric potential with respect to x and y. These partial derivatives will give us the components of the electric field intensity in the x and y directions.

Once we have the components of the electric field intensity, we can express it as a vector quantity:

E = Ex i + Ey j

Where Ex and Ey are the x and y components of the electric field intensity, respectively.

Finally, we will substitute the coordinates of the point (1, -1) into the expression for the electric field intensity to find the specific electric field intensity at that point.

This step-by-step approach will allow us to systematically solve the problem and gain a clear understanding of how to determine electric field intensity from electric potential.

To determine the electric field intensity at the point (1, -1), we will follow these steps:

Step 1: Calculate the partial derivative of V with respect to x (∂V/∂x)

Given V = -2xy, we differentiate with respect to x, treating y as a constant:

∂V/∂x = ∂(-2xy)/∂x = -2y

Step 2: Calculate the partial derivative of V with respect to y (∂V/∂y)

Similarly, we differentiate V = -2xy with respect to y, treating x as a constant:

∂V/∂y = ∂(-2xy)/∂y = -2x

Step 3: Determine the electric field intensity vector E

Using the relationship E = -∇V, we can express the electric field intensity vector as:

E = -(∂V/∂x)i - (∂V/∂y)j

Substituting the partial derivatives we calculated in steps 1 and 2, we get:

E = -(-2y)i - (-2x)j

E = 2yi + 2xj

Step 4: Evaluate E at the point (1, -1)

To find the electric field intensity at the point (1, -1), we substitute x = 1 and y = -1 into the expression for E:

E(1, -1) = 2(-1)i + 2(1)j

E(1, -1) = -2i + 2j

Therefore, the electric field intensity at the point (1, -1) is:

E = -2i + 2j N/C

This step-by-step solution demonstrates how to calculate the electric field intensity from the electric potential using the gradient operator. By following these steps, we can accurately determine the electric field intensity at any given point in space, provided we know the electric potential as a function of position.

Our calculations have revealed that the electric field intensity at the point (1, -1) is given by:

E = -2i + 2j N/C

This result is a vector quantity, meaning it has both magnitude and direction. Let's break down the components of this vector to understand its implications.

The x-component of the electric field is -2 N/C. The negative sign indicates that the electric field points in the negative x-direction. This means that a positive charge placed at the point (1, -1) would experience a force pulling it in the negative x-direction.

The y-component of the electric field is 2 N/C. The positive sign indicates that the electric field points in the positive y-direction. This means that a positive charge placed at the point (1, -1) would experience a force pushing it in the positive y-direction.

To visualize the electric field vector, imagine an arrow starting at the point (1, -1). The arrow points in the direction of the electric field, which is a combination of the negative x-direction and the positive y-direction. The length of the arrow represents the magnitude of the electric field, which can be calculated using the Pythagorean theorem:

|E| = √((-2)^2 + (2)^2) = √8 = 2√2 N/C

Therefore, the magnitude of the electric field at the point (1, -1) is 2√2 N/C.

The direction of the electric field can be described by the angle it makes with the x-axis. This angle can be calculated using the arctangent function:

θ = arctan(Ey/Ex) = arctan(2/-2) = arctan(-1) = 135°

This angle indicates that the electric field points in the second quadrant, 135 degrees from the positive x-axis.

In summary, the electric field at the point (1, -1) has a magnitude of 2√2 N/C and points in the direction 135 degrees from the positive x-axis. This information provides a complete picture of the electric force that a charge would experience at that point.

In this exploration, we successfully determined the electric field intensity at a specific point (1, -1) given the electric potential function V = -2xy. We achieved this by leveraging the fundamental relationship between electric potential and electric field intensity, which states that the electric field is the negative gradient of the electric potential.

Key takeaways from this exercise include:

• Understanding the relationship between electric potential and electric field intensity is crucial in electromagnetism.
• The electric field intensity is the negative gradient of the electric potential (E = -∇V).
• The gradient operator in Cartesian coordinates is given by ∇ = (∂/∂x)i + (∂/∂y)j + (∂/∂z)k.
• To find the electric field intensity, we need to calculate the partial derivatives of the electric potential with respect to each coordinate direction.
• The electric field intensity is a vector quantity, with both magnitude and direction.
• The components of the electric field vector indicate the direction and strength of the force that a positive charge would experience at that point.

This knowledge has wide-ranging applications in various fields, including:

• Electronics: Understanding electric fields is essential in the design and analysis of electronic circuits and devices.
• Telecommunications: Electric fields play a crucial role in the transmission and reception of electromagnetic waves in communication systems.
• Medical Imaging: Techniques like MRI and CT scans rely on the principles of electromagnetism to create images of the human body.
• Particle Physics: Electric fields are used to accelerate and manipulate charged particles in particle accelerators.
• Materials Science: The electrical properties of materials are determined by the behavior of electric fields within them.

By mastering the concepts and techniques presented in this article, you will be well-equipped to tackle a wide range of problems involving electric fields and potentials. This foundational knowledge will serve as a stepping stone for further exploration into the fascinating world of electromagnetism and its applications.

To further solidify your understanding of the concepts discussed, let's explore some practice problems:

Problem 1: The electric potential at a point in space is given by V(x, y, z) = x^2y - yz^3 + xz. Determine the electric field intensity at the point (1, 1, 1).

Solution:

1. Calculate the partial derivatives of V with respect to x, y, and z:
   • ∂V/∂x = 2xy + z
   • ∂V/∂y = x^2 - z^3
   • ∂V/∂z = -3yz^2 + x
2. Determine the electric field intensity vector E using E = -∇V:
   • E = -(2xy + z)i - (x^2 - z^3)j - (-3yz^2 + x)k
3. Evaluate E at the point (1, 1, 1):
   • E(1, 1, 1) = -(2(1)(1) + 1)i - ((1)^2 - (1)^3)j - (-3(1)(1)^2 + 1)k
   • E(1, 1, 1) = -3i - 0j + 2k

Therefore, the electric field intensity at the point (1, 1, 1) is E = -3i + 2k N/C.

Problem 2: The electric potential in a region is given by V(r) = 100/r, where r is the distance from the origin. Find the electric field intensity at a point 2 meters away from the origin.

Solution:

1. Express the gradient operator in spherical coordinates:
   • ∇ = (∂/∂r)r + (1/r)(∂/∂θ)θ + (1/(r sin θ))(∂/∂φ)φ
2. Since the potential depends only on r, the partial derivatives with respect to θ and φ are zero.
3. Calculate the partial derivative of V with respect to r:
   • ∂V/∂r = ∂(100/r)/∂r = -100/r^2
4. Determine the electric field intensity vector E using E = -∇V:
   • E = -(-100/r^2)r = (100/r^2)r
5. Evaluate E at r = 2 meters:
   • E(2) = (100/(2)^2)r = 25r

Therefore, the electric field intensity at a point 2 meters away from the origin is E = 25r N/C, where r is the unit vector in the radial direction.

These practice problems provide an opportunity to apply the concepts and techniques learned in this article. By working through these problems, you can deepen your understanding of electric potential and electric field intensity and improve your problem-solving skills.