Question

Electric field due to a point charge The electric field in the \(x y\) -plane due to a point charge at (0,0) is a gradient field with a potential function \(V(x, y)=\frac{k}{\sqrt{x^{2}+y^{2}}},\) where \(k>0\) is a physical constant. a. Find the components of the electric field in the \(x\) - and \(y\) -directions, where \(\mathbf{E}(x, y)=-\nabla V(x, y).\) b. Show that the vectors of the electric field point in the radial direction (outward from the origin) and the radial component of \(\mathbf{E}\) can be expressed as \(E_{r}=k / r^{2},\) where \(r=\sqrt{x^{2}+y^{2}}.\) c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of \(V.\)

Short answer

Question: Find the components of the electric field in the x- and y- directions and show that the electric field vectors point in the radial direction. Answer: The electric field components in the x- and y- directions are given by: \(E_x = \frac{-kx}{(x^2+y^2)^{3/2}}\) \(E_y = \frac{-ky}{(x^2+y^2)^{3/2}}\) The electric field vector in polar coordinates can be expressed as: \(\mathbf{E}(r,\theta) = -\frac{k}{r^2}(\cos{\theta}\mathbf{i} + \sin{\theta}\mathbf{j})\) Since the electric field components are expressed in polar coordinates, it is clear that the electric field vectors point in the radial direction.

Step-by-step solution

Find the gradient of the potential function

To find the components of the electric field, first, calculate the gradient of the given potential function \(V(x,y)=\frac{k}{\sqrt{x^{2}+y^{2}}}\), which can be represented as \(\nabla V(x, y)\). To find this, calculate the partial derivative of V with respect to x and y: \(E_x = \frac{\partial V}{\partial x} = \frac{-kx}{(x^2+y^2)^{3/2}}\) \(E_y = \frac{\partial V}{\partial y} = \frac{-ky}{(x^2+y^2)^{3/2}}\)

Calculate the electric field vector

Now that we have calculated the partial derivatives, we can find the components of the electric field in the x- and y- directions. Recall that \(\mathbf{E}(x, y)=-\nabla V(x, y)\), so we have: \(\mathbf{E}(x,y) = -\frac{kx}{(x^2+y^2)^{3/2}}\mathbf{i} - \frac{ky}{(x^2+y^2)^{3/2}}\mathbf{j}\)

Show that the electric field points to the radial direction and find the radial component

To show that the electric field points to the radial direction, we can rewrite the electric field vector \(\mathbf{E}(x, y)\) in polar coordinates. Let \(r = \sqrt{x^2+y^2}\). Then, we can express the electric field vector as: \(\mathbf{E}(r,\theta) = -\frac{k}{r^2}(\cos{\theta}\mathbf{i} + \sin{\theta}\mathbf{j})\) The radial component of the electric field, \(E_r\), is given by: \(E_r = \frac{k}{r^2}\)

Show that the vector field is orthogonal to the equipotential curves

Equipotential curves are defined as the level curves of the potential function where it has a constant value. The tangent to the equipotential curve at any point is determined by the gradient of the potential function at that point. If the dot product of \(\mathbf{E}(x, y)\) and \(\nabla V(x, y)\) is zero, it implies that the electric field is orthogonal to the equipotential curves. So, let's compute the dot product: \((\frac{kx}{(x^2+y^2)^{3/2}}\mathbf{i} + \frac{ky}{(x^2+y^2)^{3/2}}\mathbf{j}) \cdot (\frac{-kx}{(x^2+y^2)^{3/2}}\mathbf{i} + \frac{-ky}{(x^2+y^2)^{3/2}}\mathbf{j}) = 0\) Since the dot product is equal to zero, the vector field is orthogonal to the equipotential curves at all points in the domain of V.

Key concepts

Gradient Field

In physics, a gradient field is critical for understanding how potential changes in a space. The electric field in this context is a gradient field with a potential function. The potential function helps us calculate the electric field by determining how the potential changes in the x and y directions.

To delve into this concept, consider the potential function given, \[ V(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \]where \(k\) is a constant.

To find the gradient, calculate partial derivatives with respect to x and y:

• Partial derivative with respect to x: \[ \frac{\partial V}{\partial x} = \frac{-kx}{(x^2+y^2)^{3/2}} \]
• Partial derivative with respect to y: \[ \frac{\partial V}{\partial y} = \frac{-ky}{(x^2+y^2)^{3/2}} \]

Both components form the gradient, \( abla V(x, y)\), creating a vector field indicating the direction and magnitude of the electric force in space.

Potential Function

The potential function represents how energy is stored in the electric field due to a charge. For a point charge located at the origin, the potential function is \[ V(x, y) = \frac{k}{\sqrt{x^2+y^2}} \] where \(k > 0\).

This expression shows the potential decreases as we move further from the source (the point charge). The potential function is foundational in determining how electric fields behave because it helps visualize changes in potential energy across different positions.

To find the electric field, we use the negative gradient of this potential function:\[ \mathbf{E}(x, y) = -abla V(x, y) \] By computing the gradient we derive the electric field components along the x and y directions, enabling us to map out how forces act on other charges within this field.

Equipotential Curves

Equipotential curves are those paths or surfaces in a field where the potential remains constant. These curves represent locations where the electric potential energy is equal.

In the exercise, these curves form circles centering on the origin where the point charge is located. As you move along an equipotential curve, the potential \( V(x, y) \) remains the same, showcasing areas of equal energy.

• They are important because the electric field is always perpendicular to these curves.
• This characteristic makes it easy to find electric field lines -- just look for paths perpendicular to the equipotential curves.

Using the expression \(\mathbf{E}(x, y)\),we confirm that at any point in the electric field, the direction is orthogonal to the direction of these equipotential lines, emphasizing how energy and force orient within a field.

Radial Direction

Understanding how electric fields propagate from a point charge involves recognizing the concept of radial direction. The term ‘radial’ refers to directions radiating outwards from a central point, typically the origin where the charge is located.

The electric field derived from a point charge inherently points in the radial direction, making the electric field lines appear like rays extending from the charge: \( \mathbf{E}(x, y) = -\frac{kx}{(x^2+y^2)^{3/2}}\mathbf{i} - \frac{ky}{(x^2+y^2)^{3/2}}\mathbf{j} \).

Expressed in polar coordinates, where \( r = \sqrt{x^2+y^2} \),the radial component seen becomes:\[ E_r = \frac{k}{r^2} \]This shows that the field's intensity decreases with the square of the distance from the charge, adapting inversely with distance, and illustrating how forces diminish as they spread further away from the source.