Problem 29 - Calculus Volume 3

Question

Assume that an electric field in the \(x y\) -plane caused by an infinite line of charge along the \(x\) -axis is a gradient field with potential function \(V(x, y)=c \ln \left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right), \quad\) where \(c>0\) is a constant and \(r_{0}\) is a reference distance at which the potential is assumed to be zero. Show that the electric field at a point in the \(x y\) -plane is directed outward from the origin and has magnitude \(\mathbf{E} \mid=\frac{c}{r}, \quad\) where \(r=\sqrt{x^{2}=y^{2}}\)

Step-by-Step Solution

Answer

The electric field is directed outward from the origin and its magnitude is \( \frac{c}{r} \).

Step 1: Understand the Gradient of the Potential Function

The electric field can be found by taking the negative gradient of the potential function. Given the potential function \( V(x, y) = c \ln \left(\frac{r_0}{\sqrt{x^2 + y^2}}\right) \), we'll calculate the components separately. The gradient in the plane is\[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right). \]

Step 2: Calculate the Partial Derivative with respect to x

Calculate the partial derivative of \( V \) with respect to \( x \):\[ \frac{\partial V}{\partial x} = \frac{d}{dx}\left(c \ln \left(\frac{r_0}{\sqrt{x^2 + y^2}}\right)\right) = -\frac{c x}{x^2 + y^2}. \]

Step 3: Calculate the Partial Derivative with respect to y

Similarly, calculate the partial derivative of \( V \) with respect to \( y \):\[ \frac{\partial V}{\partial y} = \frac{d}{dy}\left(c \ln \left(\frac{r_0}{\sqrt{x^2 + y^2}}\right)\right) = -\frac{c y}{x^2 + y^2}. \]

Step 4: Express the Electric Field

The electric field \( \mathbf{E} \) is given by the negative gradient of \( V \):\[ \mathbf{E} = -abla V = \left( \frac{c x}{x^2 + y^2}, \frac{c y}{x^2 + y^2} \right). \]

Step 5: Calculate the Magnitude of the Electric Field

To find the magnitude of \( \mathbf{E} \), use the formula:\[ \|\mathbf{E}\| = \sqrt{\left(\frac{c x}{x^2 + y^2}\right)^2 + \left(\frac{c y}{x^2 + y^2}\right)^2}. \]Simplifying:\[ \|\mathbf{E}\| = \frac{c}{\sqrt{x^2 + y^2}} = \frac{c}{r}, \]where \( r = \sqrt{x^2 + y^2} \). This shows that the magnitude of the electric field is \( \frac{c}{r} \).

Step 6: Confirm the Direction of the Electric Field

The vector components \( \left( \frac{c x}{x^2 + y^2}, \frac{c y}{x^2 + y^2} \right) \) point radially outward from the origin, because they are proportional to the position vector \( (x, y) \). Thus, the direction of \( \mathbf{E} \) is outward from the origin.

Key Concepts

Potential Function

The potential function is a key concept in electrostatics, representing the potential energy per unit charge at a point in space. For our electric field caused by an infinite line charge along the x-axis, the potential function is given as:\[ V(x, y) = c \ln \left(\frac{r_{0}}{\sqrt{x^{2} + y^{2}}}\right) \]This potential function tells us how the electric potential varies with distance from the line charge. Here,

• \( c \) is a positive constant, affecting the strength of the potential.
• \( r_{0} \) is a reference distance, where the potential is zero.

The function uses a logarithm, indicating that the potential decreases with the distance \( \sqrt{x^2 + y^2} \) from the line, capturing how the influence of charge spreads out in space.

Gradient Field

A gradient field, in physics, is created from a scalar potential function and represents a vector field showing the direction and rate of change of that function. In the context of the potential function \( V(x, y) \), the gradient field is the electric field \( \mathbf{E} \), and it's calculated using:\[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) \]The gradient points in the direction of the greatest rate of increase of the potential function, and its magnitude indicates how steep that increase is. The fact that the electric field is a gradient field implies that it can be derived entirely from the potential function, highlighting the close relationship between potential and electric fields.

Partial Derivative

Partial derivatives are used to find how a function changes as each independent variable changes, while keeping the others constant. For our potential function, we need to find these derivatives with respect to both \( x \) and \( y \):

• For \( x \): \[ \frac{\partial V}{\partial x} = -\frac{c x}{x^2 + y^2} \]
• For \( y \): \[ \frac{\partial V}{\partial y} = -\frac{c y}{x^2 + y^2} \]

Partial derivatives allow us to examine the local behavior of potential at each point in the \( xy \)-plane. Calculating these helps us find how both the potential and consequently the electric field change as you move in space.

Magnitude of Electric Field

The magnitude of the electric field \( \mathbf{E} \) is derived from the components of the electric field vector, which are the negative of the gradients of the potential function with respect to the coordinates \( x \) and \( y \). Thus, the vector \( \mathbf{E} \) is:\[ \mathbf{E} = \left( \frac{c x}{x^2 + y^2}, \frac{c y}{x^2 + y^2} \right) \]To find the magnitude of the electric field, use the formula:\[ \| \mathbf{E} \| = \sqrt{\left(\frac{c x}{x^2 + y^2}\right)^2 + \left(\frac{c y}{x^2 + y^2}\right)^2} \]This calculation simplifies to:\[ \| \mathbf{E} \| = \frac{c}{r}, \]where \( r = \sqrt{x^2 + y^2} \), the distance from the origin. The magnitude \( \frac{c}{r} \) shows how the electric field's strength weakens as you move further from the origin, consistent with the properties of a radial field extending outward from a line charge.