Problem 97 - University Physics with Modern Physics

Question

Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x = \frac{1}{2} a\) and \(x = \frac{1}{2} a + L\) and the other between \(x = -\frac{1}{2} a\) and \(x = -\frac{1}{2} a - L\). Each rod has positive charge \(Q\) distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive x-axis. (b) Show that the magnitude of the force that one rod exerts on the other is $$F = {Q^2 \over 4\pi\epsilon_0 L^2} ln [ {(a + L)^2 \over a(a + 2L)} ]$$ (c) Show that if \(a\) \(\gg\) \(L\), the magnitude of this force reduces to \(F = Q^2/4\pi\epsilon_0 a^2\). (\(Hint\): Use the expansion ln \((1 + z) = z - \frac{1}{2} z^2 + \frac{1}{3} z^3 - \cdot\cdot\cdot\), valid for \(\mid z \mid\ll1\). Carry \(all\) expansions to at least order \(L^2/a^2.\)) Interpret this result.

Step-by-Step Solution

Answer

(a) Electric field expression: \( E = \frac{Q}{4\pi\epsilon_0 L} \left( \frac{1}{x + \frac{a}{2}} - \frac{1}{x + \frac{a}{2} + L} \right) \). (b) Force magnitude: \( F = \frac{Q^2}{4\pi\epsilon_0 L^2} \ln \left( \frac{(a + L)^2}{a(a + 2L)} \right) \). (c) For \( a \gg L \), \( F \approx \frac{Q^2}{4\pi\epsilon_0 a^2} \).

Step 1: Expression for electric field from a differential element

Consider a small element of charge, \( dq \), on the second rod located at position \( x \) such that \( x = -\frac{a}{2} - x' \) where \( 0 \leq x' \leq L \). The charge element can be expressed as \( dq = \frac{Q}{L}dx' \). The electric field generated by this small charge at a point \( P \) on the positive \( x \)-axis a distance \( x \) from the origin is given by Coulomb's law: \[ dE = \frac{k \cdot dq}{r^2} \] where \( r \) is the distance from the charge element to point \( P \) and \( k = \frac{1}{4\pi\epsilon_0} \).

Step 2: Calculate the net electric field on the positive x-axis

The distance from \( dq \) to a point on the positive \( x \)-axis at position \( x \) is \( r = x + \frac{a}{2} + x' \). Therefore, the differential electric field is: \[ dE = \frac{\frac{Q}{L}\,dx'}{4\pi\epsilon_0 \left(x + \frac{a}{2} + x'\right)^2} \] The direction of the electric field is along the x-direction (positive x-axis). Therefore, the net electric field \( E \) is obtained by integrating \( dE \) from \( x' = 0 \) to \( x' = L \).\[ E = \int_0^L \frac{\frac{Q}{L}}{4\pi\epsilon_0 \left(x + \frac{a}{2} + x'\right)^2} \, dx' \]

Step 3: Solve the integral to find the electric field

To solve the integral, perform the integration:\[ E = \frac{Q}{4\pi\epsilon_0 L} \int_0^L \frac{1}{\left(x + \frac{a}{2} + x'\right)^2} \, dx' \] Using a simple substitution \( u = x + \frac{a}{2} + x' \), \( du = dx' \), the limits change from \( x' = 0 \) to \( x' = L \) become \( u = x + \frac{a}{2} \) to \( u = x + \frac{a}{2} + L \).Therefore, the integral becomes:\[ E = \frac{Q}{4\pi\epsilon_0 L} \left[ -\frac{1}{u} \right]_{x + \frac{a}{2}}^{x + \frac{a}{2} + L} \]Which simplifies to:\[ E = \frac{Q}{4\pi\epsilon_0 L} \left(\frac{1}{x + \frac{a}{2}} - \frac{1}{x + \frac{a}{2} + L}\right) \]

Step 4: Calculate the force and derive result for force between rods

The electric field from the second rod exerts a force on the first rod. The differential force is given by:\[ dF = dq \cdot E \]where the charge element \( dq = \frac{Q}{L} \, dx \) along the first rod. Thus,\[ dF = \frac{Q^2}{4\pi\epsilon_0 L^2} \cdot \left(\frac{1}{x + \frac{a}{2}} - \frac{1}{x + \frac{a}{2} + L}\right) dx \]Integrate \( dF \) over the length of the first rod from \( x = \frac{a}{2} \) to \( x = \frac{a}{2} + L \):\[ F = \frac{Q^2}{4\pi\epsilon_0 L^2} \int_{\frac{a}{2}}^{\frac{a}{2} + L} \left(\frac{1}{x + \frac{a}{2}} - \frac{1}{x + \frac{a}{2} + L}\right) dx \]This evaluates to:\[ F = \frac{Q^2}{4\pi\epsilon_0 L^2} \ln \left( \frac{(a + L)^2}{a (a + 2L)} \right) \]

Step 5: Simplify force for condition when \( a \gg L \)

When \( a \gg L \), we can use the expansion \( \ln(1+z) \approx z - \frac{1}{2}z^2 \) for small \( z \). Set \( z = \frac{2L}{a} \) for \( \frac{a + 2L}{a + L} \approx 1 + \frac{2L}{a} \). This leads to expanded terms:\[ \ln \left( \frac{(a+L)^2}{a(a+2L)} \right) \approx \ln \left( 1 + \frac{L^2}{a^2} \right) \approx \frac{L^2}{a^2} \] Plugging \( \ln \left( 1 + \frac{L^2}{a^2} \right) \) into the force equation:\[ F \approx \frac{Q^2}{4\pi\epsilon_0 a^2} \]which is the simplified result for \( a \gg L \).

Step 6: Interpretation of Result

The derived expression for force \( F \approx \frac{Q^2}{4\pi\epsilon_0 a^2} \) is the standard expression for the force between two point charges separated by distance \( a \). This result indicates that when the rods are far apart compared to their lengths, the rods behave like point charges located at their centers.

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding electric interactions between charged objects. It describes the force between two point charges. According to this law, the electric force ( \( F_e \) ) between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as:

• \( F_e = \frac{{k \, |q_1 \, q_2|}}{{r^2}} \)
• Where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance separating the charges.

This formula is crucial when calculating the electric field created by a charged object at a certain point. In our example, each rod is considered to be made of numerous small point charges. The collective effect of these can be integrated to find the total electric field they produce.

Electric Force

Electric force is the influence exerted by an electric field on a charged object. It can move or change the direction of the movement of the electrical charges. The force itself is what is calculated using Coulomb's Law.In the context of the rods mentioned in our problem, the electric field from one rod exerts a force on the other. The force can be understood as the sum of the interactions of small charge segments across the entire length of the rods.In this problem, the expression:

• \[ F = \frac{Q^2}{4\pi\epsilon_0L^2} \ln \left( \frac{(a + L)^2}{a(a + 2L)} \right) \]

represents the total force exerted by one rod on another. This formula results from integrating all the differential forces (termed \( dF \)) along the rod due to the electric field.

Integration

Integration is a powerful mathematical tool used to compute quantities when dealing with continuous charge distributions. In the context of electric fields, integration helps to sum up contributions from infinitesimally small charge elements to find the total electric field or force.When you calculate the electric field from a rod, each small charge element contributes a small electric field. To find the total field, you integrate these small contributions over the length of the rod. This process transforms a seemingly complex problem of many individual electric forces into a single equation:

• \[ E = \int \limits_{0}^{L} \frac{\frac{Q}{L}}{4\pi\epsilon_0 \left(x + \frac{a}{2} + x'\right)^2} \, dx' \]

This is an integration with respect to the position along the rod, denoted as \( x' \). Each point \( x' \) gives a different contribution to the field at a specific point in space.

Uniform Charge Distribution

Uniform charge distribution is a term used when charge is spread evenly across an object, such as a rod. When each part of a rod carries an equal amount of charge per unit length, it is termed to have a uniform charge distribution.In our scenario, both rods have a uniform distribution represented by a charge density \( \lambda \), defined as:

• \( \lambda = \frac{Q}{L} \)

where \( Q \) is the total charge and \( L \) is the length of the rod.This uniform distribution simplifies calculations. It allows us to use integration effectively to determine fields and forces. It also ensures consistency in calculations, as each segment of the rod is treated equally in terms of charge contribution.