Question

Let \(V=2 x y^{2} z^{3}+3 \ln \left(x^{2}+2 y^{2}+3 z^{2}\right) \mathrm{V}\) in free space. Evaluate each of the following quantities at \(P(3,2,-1)(a) V ;(b)|V| ;(c) \mathbf{E} ;(d)|\mathbf{E}| ;(e) \mathbf{a}_{N}\); (f) D.

Short answer

Question: Determine the following quantities regarding the potential function \(V(x, y, z) = 2xyz^3 + 3\ln(x^2 + 2y^2 + 3z^2)\) at the point P(3,2,-1). (a) The value of V at the point P. (b) The absolute value of V at the point P. (c) The electric field E at the point P. (d) The absolute value of the electric field E at the point P. (e) The normal acceleration aN at the point P. (f) The electric flux density D at the point P. Answer: (a) The value of V at the point P is \(V(3,2,-1) = -24 + 3\ln[13]\). (b) The absolute value of V at the point P is \(|V(3,2,-1)| = | -24 + 3\ln[13] |\). (c) The electric field E at point P is \(\mathbf{E}(3,2,-1) = -(-16 + \frac{18}{13}\mathbf{i} - 24 + \frac{24}{13}\mathbf{j} + 72 - \frac{18}{13}\mathbf{k})\). (d) The absolute value of the electric field E at point P is \(|\mathbf{E}(3,2,-1)| = \sqrt{\left(-16 + \frac{18}{13}\right)^2 + \left(-24 + \frac{24}{13}\right)^2 + \left(72 - \frac{18}{13}\right)^2}\). (e) There is not enough information given to determine the normal acceleration aN at point P. (f) The electric flux density D at point P is \(\mathbf{D}(3, 2, -1) = 8.854 \times 10^{-12}(-(16 - \frac{18}{13}\mathbf{i} - 24 + \frac{24}{13}\mathbf{j} + 72 - \frac{18}{13}\mathbf{k}))\).

Step-by-step solution

Find the value of V at point P(3,2,-1)

To find the value of the potential function V at the point (3, 2, -1), substitute the coordinates of this point into the function: \(V(3,2,-1) = 2(3)(2^2)(-1)^3 + 3\ln[(3^2 + 2(2^2) + 3(-1)^2)]\) Calculate the resulting value: \(V(3,2,-1) = 2(3)(4)(-1) + 3\ln[13] = -24 + 3\ln[13]\)

Find the absolute value of V at point P

To find the absolute value of V at point P, compute the absolute value of the expression found above: \(|V(3,2,-1)| = | -24 + 3\ln[13] |\)

Find the gradient of V

The electric field E can be found by calculating the gradient of V: \(\mathbf{E} = -\nabla V = -\frac{\partial V}{\partial x}\mathbf{i} -\frac{\partial V}{\partial y}\mathbf{j} -\frac{\partial V}{\partial z}\mathbf{k}\) Calculate the partial derivatives of V: \(\frac{\partial V}{\partial x} = 2y^2 z^3 + 3\frac{2x}{x^2 + 2y^2 + 3z^2}\) \(\frac{\partial V}{\partial y} = 4xyz^3 + 3\frac{4y}{x^2 + 2y^2 + 3z^2}\) \(\frac{\partial V}{\partial z} = 6x y^2 z^2 + 3\frac{6z}{x^2 + 2y^2 + 3z^2}\)

Find the electric field E at point P

Substitute the coordinates of P into the partial derivatives of V and then use the gradient formula to find E: \(\mathbf{E}(3,2,-1) = -(\frac{\partial V}{\partial x}\mathbf{i} +\frac{\partial V}{\partial y}\mathbf{j} +\frac{\partial V}{\partial z}\mathbf{k})\) at P Calculate the resulting electric field E: \(\mathbf{E}(3,2,-1) = -(2(2^2)(-1)^3 + 3\frac{6}{13}\mathbf{i} + 4(3)(2)(-1)^3 + 3\frac{8}{13}\mathbf{j} + 6(3)(2^2)(-1)^2 + 3\frac{-6}{13}\mathbf{k})\) \(\mathbf{E}(3,2,-1) = -(-16 + \frac{18}{13}\mathbf{i} - 24 + \frac{24}{13}\mathbf{j} + 72 - \frac{18}{13}\mathbf{k})\)

Find the absolute value of the electric field E at point P

To find the absolute value of the electric field E at point P, compute the magnitude of the vector obtained above: \(|\mathbf{E}(3,2,-1)| = \sqrt{\left(-16 + \frac{18}{13}\right)^2 + \left(-24 + \frac{24}{13}\right)^2 + \left(72 - \frac{18}{13}\right)^2}\)

Find the normal acceleration aN at point P

We have to use chain rule to find the normal acceleration aN. However, the given problem does not provide enough information (no mention of time dependence or motion) to determine the normal acceleration aN at point P.

Find the electric flux density D at point P

In free space, D is related to E by the formula: \(\mathbf{D} = \epsilon_0 \mathbf{E}\) Use the value of permittivity of free space (\(\epsilon_0 = 8.854 \times 10^{-12} F/m\)) and the electric field obtained earlier to calculate the electric flux density at P: \(\mathbf{D}(3, 2, -1) = 8.854 \times 10^{-12}\mathbf{E}(3, 2, -1)\) Calculate the resulting electric flux density D: \(\mathbf{D}(3, 2, -1) = 8.854 \times 10^{-12}(-(16 - \frac{18}{13}\mathbf{i} - 24 + \frac{24}{13}\mathbf{j} + 72 - \frac{18}{13}\mathbf{k}))\)

Key concepts

Electrostatic Potential (V)

Electrostatic potential, denoted by the symbol 'V', represents the potential energy per unit charge at a specific point in an electric field. It is a scalar quantity, meaning it has magnitude but no direction. The potential at a point due to an electric charge is equivalent to the work done in bringing a unit positive charge from infinity to that point.

For a more practical understanding, consider the formula given in the exercise:
\(V = 2 x y^{2} z^{3} + 3 \ln(x^{2} + 2 y^{2} + 3 z^{2}) V\).
This expression is a function of the coordinates 'x', 'y', and 'z', indicating that the electrostatic potential varies with position in space. To find the potential at a particular point, such as P(3,2,-1), we plug these values into the potential function and simplify the resulting equation to obtain the potential 'V' at point P. This is an essential concept in electromagnetics as it helps to understand the behavior of electric fields and the forces they exert on charges.

Electric Field Intensity (E)

Electric field intensity, often symbolized by 'E', is a vector quantity that represents the force per unit charge exerted on a positive test charge placed within the field. The direction of the electric field is the direction in which a positive test charge would move if placed within the field.

To find 'E' at a certain point, one must take the gradient of the electrostatic potential 'V'. The gradient operation (covered in a later section) yields not only the magnitude but also the direction of the field. Mathematically, the electric field \(\mathbf{E} = -abla V\)
is negative because it points in the direction of decreasing potential. By applying the gradient to 'V', we obtain the components of the electric field at point P, demonstrating how 'V' relates to 'E' in a spatially varying electric field.

Electric Flux Density (D)

Electric flux density, designated as 'D', is a measure of the electric field's distribution over a given area. For a simple understanding, it can be seen as how 'spread out' the electric field lines are. In free space, the electric flux density is directly proportional to the electric field intensity.

The constant of proportionality is the permittivity of free space, denoted as \(\epsilon_0\).
The relationship between 'D' and 'E' is given by the equation \(\mathbf{D} = \epsilon_0 \mathbf{E}\).
By using the electric field intensity calculated earlier, one can find the electric flux density at the same point. This concept is vital in electromagnetics for understanding how electric fields interact with materials and in designing capacitors and other electrical components.

Gradient Operation in Electromagnetics

The gradient operation is a mathematical tool used to determine the rate and direction of change in a scalar field. In electromagnetics, the gradient of the electrostatic potential 'V' gives us the electric field intensity 'E'.

Mathematically, the gradient is represented as \(abla V\),
and when calculated, it provides the three components of the electric field corresponding to the 'x', 'y', and 'z' directions. Performing the gradient operation for our given potential function 'V' at any point gives us a vector that points in the direction of the greatest rate of increase of the potential and whose magnitude is the rate of that increase per unit distance.

Absolute Value of Electric Quantities

The absolute value of electric quantities, such as the electrostatic potential 'V' or the electric field intensity 'E', refers to their magnitude regardless of their direction or sign. Taking the absolute value is akin to measuring how strong these quantities are without considering their vector nature.

For potential 'V', the absolute value is found by taking the modulus of the calculated potential at point P, which in mathematical terms is represented as \(|V|\).
Similarly, the magnitude of the electric field is found by computing the Euclidean norm of the vector \(|\mathbf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2}\),
where \(E_x\), \(E_y\), and \(E_z\) are the components of 'E'. This gives us a single, non-negative number indicating the strength of the electric field at point P, regardless of its direction.