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Over a certain region of space, the electric potential is \(V=5 x-3 x^{2} y+2 y z^{2} .\) Find the expressions for the \(x, y\) and \(z\) components of the electric field over this region. What is the magnitude of the field at the point \(P\) that has coordinates (1,0,-2) m?

Short Answer

Expert verified
The components of the electric field are \(E_x = -5 + 6xy\), \(E_y = 3x^2 - 4yz^2\), and \(E_z = -4yz\). At the point P(1,0,-2), the magnitude of the electric field is 5 N/C.

Step by step solution

01

Finding the x-component

The x-component of the electric field can be found by deriving the potential with respect to \(x\) and then change the sign: \(-\frac{\partial V}{\partial x}\). Using the derivative results in the expression for the x component: \(E_x = -5 + 6xy\).
02

Finding the y-component

The y-component can be found by similar means as the x-component, but this time the potential is derived with respect to \(y\), resulting in: \(E_y = 3x^2 - 4yz^2\).
03

Finding the z-component

The z-component can be found in a similar way by deriving the potential with respect to \(z\): \(E_z = -4yz\).
04

Calculating the magnitude

If the components of the electric field are known, the magnitude can be calculated using the Pythagorean theorem in three dimensions: \(E = \sqrt{E_x^2 + E_y^2 + E_z^2}\).
05

Figure out the magnitude at point P

To find the magnitude at point P (1,0,-2), substitute the coordinates into the expressions for each component: \(E_x = -5\), \(E_y = 0\), and \(E_z = 0\). Substituting these values into the formula for the field magnitude gives: \(E = 5\) N/C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential
Electric potential is a fundamental concept in electromagnetism, representing the potential energy per unit charge at a given point in space. It is typically denoted by the symbol \( V \) and is measured in volts. The electric potential allows us to understand how much work is needed to move a charge within an electric field, or conversely, how much work it can perform. In the given exercise, the electric potential is expressed with the formula \( V = 5x - 3x^2y + 2yz^2 \). This formula gives us the potential at any point in space defined by coordinates \( (x, y, z) \). From this expression, we can derive the electric field, which shows how the potential changes within the region.
Field Components
The electric field is a vector field represented by components in the \( x \), \( y \), and \( z \) directions. These components describe how the field behaves spatially in each direction. To find each component, we derive the electric potential function with respect to the respective coordinates:
  • **\( x \)-component**: Found by partially deriving \( V \) with respect to \( x \), yielding \( E_x = -\frac{\partial V}{\partial x} \).
  • **\( y \)-component**: Obtained by partially deriving \( V \) with respect to \( y \), providing \( E_y = -\frac{\partial V}{\partial y} \).
  • **\( z \)-component**: Derived by taking \( V \)'s partial derivative with respect to \( z \), resulting in \( E_z = -\frac{\partial V}{\partial z} \).
Each negative sign in the component equations reflects that the electric field points in the direction of decreasing potential.
Magnitude of Electric Field
The magnitude of the electric field represents the strength of the field at a particular point and is calculated using the Pythagorean theorem for three dimensions. Once you have the expressions for each component \( E_x \), \( E_y \), and \( E_z \), the magnitude \( E \) is calculated as follows: \[E = \sqrt{E_x^2 + E_y^2 + E_z^2} \]This method provides a scalar value that indicates how intense the electric field is at any location in space. Specifically, at the given point \((1,0,-2)\), the field's magnitude was found to be \( 5 \text{ N/C} \), derived from substituting the field component values into this equation.
Partial Derivatives
Partial derivatives are a crucial mathematical tool used to find how a multivariable function changes with respect to one of those variables, holding the others constant. In the context of electric fields and potentials, they help us determine the rate of change of the potential in different spatial directions. For instance, when calculating the electric field components:
  • **With respect to \( x \)**: Results in \( \frac{\partial V}{\partial x} = 5 - 6xy \).
  • **With respect to \( y \)**: Yields \( \frac{\partial V}{\partial y} = -3x^2 + 4yz^2 \).
  • **With respect to \( z \)**: Gives \( \frac{\partial V}{\partial z} = 4yz \).
These derivatives are directly used to calculate the electric field components. Understanding partial derivatives simplifies the process of analyzing changes in the electric potential and facilitates the step-by-step determination of each electric field vector component.

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Most popular questions from this chapter

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