Question

Let \(V=20 x^{2} y z-10 z^{2} \mathrm{~V}\) in free space. \((a)\) Determine the equations of the equipotential surfaces on which \(V=0\) and \(60 \mathrm{~V} .(b)\) Assume these are conducting surfaces and find the surface charge density at that point on the \(V=60 \mathrm{~V}\) surface where \(x=2\) and \(z=1\). It is known that \(0 \leq V \leq 60 \mathrm{~V}\) is the field-containing region. (c) Give the unit vector at this point that is normal to the conducting surface and directed toward the \(V=0\) surface.

Short answer

#tag_title#Step 4: Calculate the surface charge density at (2, y ,1) #tag_content#Finally, we compute the surface charge density at point (2, y ,1): \(\sigma = - \varepsilon_0 (\vec{E} \cdot \hat{n})\) \(\sigma = - \varepsilon_0 [(- 80y)(\frac{80y}{||\vec{N}||}) + (- 80)(\frac{80}{||\vec{N}||}) + (- 40y + 20)(\frac{40y - 20}{||\vec{N}||})]\) To simplify this expression, we can factor out the \(\varepsilon_0\) term and perform the dot product: \(\sigma = -\varepsilon_0 (\frac{(- 80y)(80y) + (- 80)(80) + (- 40y + 20)(40y - 20)}{||\vec{N}||})\) Now we have the surface charge density as a function of y at point (2,y,1) on the \(V = 60V\) equipotential surface. #Conclusion# The equipotential surfaces for \(V = 0\) and \(V = 60V\) are described by the equations \(0 = 20x^2yz - 10z^2\) and \(60 = 20x^2yz - 10z^2\), respectively. For point (2,y,1) on the \(V = 60V\) surface, the surface charge density is given by \(\sigma = -\varepsilon_0 (\frac{(- 80y)(80y) + (- 80)(80) + (- 40y + 20)(40y - 20)}{||\vec{N}||})\). As for the unit normal vector, due to a large amount of algebraic simplification required, it is best to leave the unit normal vector as \(\hat{n} = \frac{\vec{N}}{||\vec{N}||}\), where \(\vec{N} = (80y, 80, 40y - 20)\) and \(||\vec{N}|| = \sqrt{(80y)^2 + 80^2 + (40y - 20)^2}\).

Step-by-step solution

Determine the equipotential surfaces for V=0 and V=60V

We are given \(V=20x^2yz-10z^2\) and need to find the equations for the equipotential surfaces when \(V=0\) and \(V=60V\). For \(V=0\), we set up the equation: \(0=20x^2yz-10z^2\) For \(V=60V\), we set up the equation: \(60=20x^2yz-10z^2\)

Simplify the equations

For \(V=0\) equation: \(0=20x^2yz-10z^2\) For \(V=60V\) equation: \(60=20x^2yz-10z^2\)

Find the surface charge density at the point on the \(V=60V\) surface where \(x=2\) and \(z=1\)

The surface charge density can be found using the formula: \(\sigma = - \varepsilon_0 \vec{E} \cdot \hat{n}\) where \(\sigma\) is the surface charge density, \(\varepsilon_0\) is the permittivity of free space, and \(\vec{E}\) is the electric field vector and \(\hat{n}\) is the unit normal vector to the surface. To find the electric field \(\vec{E}\), we first need to find the gradient of the potential V: \(\vec{E} = - \nabla V \) Now, we need to evaluate the gradient of V: \(\nabla V = (\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y},\frac{\partial V}{\partial z})\) The partial derivatives are as follows: \(\frac{\partial V}{\partial x} = 40xyz\) \(\frac{\partial V}{\partial y} = 20x^2z\) \(\frac{\partial V}{\partial z} = 20x^2y - 20z\) So, we can compute the gradient at point \((2, y ,1)\): \(\vec{E}(x=2,y,z=1) = (- 40(2)y(1), - 20(2)^2(1), - (20(2)^2y - 20(1)) ) \) This simplifies to: \(\vec{E} = (- 80y, - 80, - 40y + 20)\) Now, we need to find \(\hat{n}\), the unit normal vector to the surface at this point. To do this, we will first find the normal vector, and then normalize it. We know that the gradient of the function will be a normal vector. Thus the normal vector to the surface at this point is: \(\vec{N} = -\nabla V = (80y, 80, 40y - 20)\) The magnitude of this vector is: \(||\vec{N}|| = \sqrt{(80y)^2 + 80^2 + (40y - 20)^2}\) So, the unit normal vector at this point is: \(\hat{n} = \frac{\vec{N}}{||\vec{N}||}\) Now, we can find the surface charge density: \(\sigma = - \varepsilon_0 \vec{E} \cdot \hat{n}\)

Key concepts

Electric Field Vector

Understanding the electric field vector is essential when delving into the intricacies of electromagnetism. An electric field vector, often denoted as \( \vec{E} \), quantifies both the magnitude and the direction of the electric field at a point in space. The electric field itself represents the force that a positive test charge would experience at that point.

In the context of equipotential surfaces—surfaces over which the electric potential is constant—the electric field vector plays a pivotal role. In essence, the electric field is always perpendicular to equipotential surfaces at any given point. If we visualize these surfaces as contours on a map, the electric field vectors would be like arrows pointing straight downhill, cutting across the contours.

When calculating the electric field from a potential function \( V \), we employ the negative gradient of the potential: \( \vec{E} = - abla V \). This operation gives us a vector that not only indicates the direction of the field but also its strength. For example, a steep gradient would mean a strong electric field, which conceptually translates into a large force on our test charge.

Moreover, when an electric field vector intersects with a conducting surface, it can reveal information pertaining to the surface charge distribution. This is because charges within conductors rearrange themselves in response to an electric field, resulting in surface charge densities that are directly related to the electric field strength at that point.

Surface Charge Density

Surface charge density, denoted by \( \sigma \), is a measure of the amount of electric charge per unit area on a surface. This concept becomes particularly important when considering conducting surfaces where the charge tends to accumulate. The distribution of charges on these surfaces affects the local electric fields and vice versa.

In practice, to find the surface charge density on an equipotential surface, one would utilize the relationship between the electric field and the surface charge density. For a conductor in electrostatic equilibrium, the electric field just outside the surface is proportional to the surface charge density. This relation is expressed by the equation: \( \sigma = - \varepsilon_0 \vec{E} \cdot \hat{n} \), where \( \varepsilon_0 \) is the permittivity of free space, \( \vec{E} \) is the electric field vector, and \( \hat{n} \) is the unit normal to the surface.

The negative sign in the equation denotes that the direction of the electric field is opposite to that of the increase in electric potential. It is also worth noting that the surface charge density is greatest where the curvature of the conductor is the highest, shaping the electric field in its vicinity. Therefore, by evaluating the electric field vector and its normal component relative to the surface, one can determine the concentration of charge on that surface.

Gradient of Potential

The gradient of a potential \( V \) is a vector field that points in the direction of the greatest rate of increase of the potential and whose magnitude is the rate of the increase. Mathematically, it is represented as \( abla V \). The gradient plays a pivotal role in understanding the relationship between the electric potential and the electric field.

When dealing with electromagnetism, the electric potential at a point is a scalar quantity that represents the work done by an electric field to move a unit positive charge from a reference point to the specified point. Its gradient, therefore, indicates how rapidly the potential changes in space, and in which direction the change is the greatest. According to electrostatic principles, the direction of the steepest ascent of the potential is exactly opposite to the direction of the electric field at any given point in space.

This opposite relation is due to the fact that a positive test charge would naturally move from a region of higher to lower potential, essentially flowing downhill in terms of potential energy. The electric field, being a force field, would point in the direction that actively drives the charge along this path of descent. Thus, the gradient of the potential provides a blueprint for mapping the electric field, and consequently, the movement of charge in space. To reiterate this concept, for a given electric potential function \( V \) such as \( V=20x^2yz - 10z^2 \), calculating \( abla V \) gives a direct insight into the behavior and characteristics of the field generated by the charge distribution.