Problem 20 - University Physics with Modern Physics

Question

A positive charge \(+q\) is located at the point \(x=0\) \(y=-a,\) and a negative charge \(-q\) is located at the point \(x=0,\) \(y=+a .(\) a) Derive an expression for the potential \(V\) at points on the \(y\) -axis as a function of the coordinate \(y .\) Take \(V\) to be zero at an infinite distance from the charges. (b) Graph \(V\) at points on the \(y\) -axis as a function of \(y\) over the range from \(y=-4 a\) to \(y=+4 a\) (c) Show that for \(y>a\) , the potential at a point on the positive \(y\) -axis is given by \(V=-\left(1 / 4 \pi \epsilon_{0}\right) 2 q a / y^{2}\) . (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a ?\)

Step-by-Step Solution

Answer

(a) \(V(y) = \frac{kq}{y+a} - \frac{kq}{y-a}\). (c) For \(y>a\), \(V\approx -\frac{2kaq}{y^2}\). Interchanging charges does not change the potential's functional form.

Step 1: Understand the Charges Configuration

We have two point charges, a positive charge \(+q\) at point \((0, -a)\) and a negative charge \(-q\) at point \((0, +a)\). We want to find the electric potential \(V\) at any point on the y-axis due to these charges.

Step 2: Write the Expression for Potential Due to Each Charge

The electric potential \(V\) due to a point charge \(q\) at a distance \(r\) is given by \(V = \frac{kq}{r}\), where \(k = \frac{1}{4\pi \epsilon_0}\). For a point on the y-axis at \((0, y)\), the distance to the positive charge is \(|y + a|\) and to the negative charge is \(|y - a|\).

Step 3: Derive the Expression for Total Potential

Using the principle of superposition, the total potential \(V(y)\) on the y-axis is the sum of potentials due to the positive and negative charges:\[V(y) = \frac{kq}{|y + a|} - \frac{kq}{|y - a|}.\]

Step 4: Graph the Potential Over the Range

To visualize \(V(y)\), plot \(V\) as a function of \(y\) over the interval \([-4a, 4a]\). Note that the potential is zero at \(y = \pm a\) and will show symmetry about these points due to the configuration of charges.

Step 5: Simplify for Points Where \(y > a\)

For points where \(y > a\), the absolute distance simplifies to regular expressions:\[V(y) = \frac{kq}{y + a} - \frac{kq}{y - a}\approx \frac{-2kaq}{y^2}.\]This approximation uses binomial expansion assuming \(a \ll y\).

Step 6: Consider Interchanged Charges

If the charges are interchanged, the positions of positive and negative charges switch, reversing their influence on the potential expression derived in step 3. The new total potential will still be:\[V(y) = -\left( \frac{1}{4\pi \epsilon_0} \right) \frac{2qa}{y^2},\]but it affects the sign depending on distance from each charge.

Key Concepts

Electrostatics

Electrostatics is the branch of physics that deals with the study of forces, fields, and potentials arising from static or stationary electric charges. When charges are stationary, they create an electric field and potential around them.
Electric potential, often denoted by the symbol \(V\), is a measure of potential energy per unit charge at a specific point in an electric field. It can be thought of as how much work would be required to bring a unit positive charge from infinity to that point without any acceleration.
Electric fields are responsible for exerting forces on charges, leading to potential differences, which are crucial in understanding how electrostatic interactions occur. Understanding electrostatics is fundamental to comprehending many concepts in electricity and magnetism, as it forms the basis for how charges interact in various environments.

Point Charges

Point charges are idealized charges that are assumed to exist at a single point in space. These charges are often used in physics problems to simplify the mathematical treatment of electric fields and potentials.
A point charge creates an electric field radiating outward (or inward if the charge is negative) and decreases in magnitude with distance. The electric potential due to a point charge is higher closer to the charge and decreases as we move away.

• The potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by the formula \( V = \frac{kq}{r} \), where \( k = \frac{1}{4\pi \epsilon_0} \).
• This formula shows that the potential depends inversely on the distance, meaning the farther away you are, the less influence the charge has.

Understanding point charges is essential, as they serve as the basic building blocks for more complex charge distributions and are involved in many theoretical analyses.

Superposition Principle

The superposition principle states that the total electric potential at any point due to a number of point charges is simply the algebraic sum of the potentials due to each charge taken separately.
This principle is powerful because it allows us to analyze systems with multiple charges effectively, by considering each charge's influence independently and then summing them up.
For example, if you're trying to find the electric potential at a point due to multiple charges, calculate the potential from each charge at that point, and then sum them up:

• For the exercise, \( V(y) = \frac{kq}{|y + a|} - \frac{kq}{|y - a|} \), where each term represents the potential due to the respective charge.

The superposition principle is a cornerstone in electrostatics and is applicable not only to potentials but also to electric fields, allowing simplification of complex charge configurations.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the electrostatic force \( F \) between any two charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the charges and inversely proportional to the square of the distance \( r \) between them:
\[ F = k \frac{q_1 q_2}{r^2} \]This ensures that the force decreases rapidly as the distance increases, which is why you often see a notable influence of charges only when they are relatively close.

• The constant \( k \) is the Coulomb's constant \( k = \frac{1}{4\pi \epsilon_0} \).
• This law is fundamental in analyzing the forces between charges and helps in determining potential energy and electric field direction and magnitude.

In solving potential problems, such as the one described, understanding Coulomb's Law is crucial because it helps to derive the electric field, which in turn can be used to find the potential at different points in the field.