Two thin rods, each with length L and total charge +Q, are...

Two thin rods, each with length $L$ and total charge $+Q,$ are parallel and separated by a distance $a .$ The first rod has one end at the origin and its other end on the positive $y$ -axis. The second rod has its lower end on the positive $x$ -axis. (a) Explain why the $y$ -component of the net force on the second rod vanishes. (b) Determine the $x$ -component of the differential force $d F_{2}$ exerted on a small portion of the second rod, with length $d y_{2}$ and position $y_{2},$ by the first rod. (This requires integrating over differential portions of the first rod, parameterized by $\left.d y_{1} .\right)$ (c) Determine the net force $\vec{F}_{2}$ on the second rod by integrating $d F_{2 x}$ over the second rod. (d) Show that in the limit $a \gg L$ the force determined in part (c) becomes $\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{a^{2}} \hat{\imath}$. (e) Determine the external work required to move the second rod from very far away to the position $x=a$, provided the first rod is held fixed at $x=0 .$ This describes the potential energy of the original configuration. (f) Suppose $L=50.0 \mathrm{~cm}, a=10.0 \mathrm{~cm}, Q=10.0 \mu \mathrm{C},$ and $m=500 \mathrm{~g}$. If the two rods are released from the original configuration, they will fly apart and ultimately achieve a particular relative speed. What is that relative speed?

Two thin rods, each with length $L$ and total charge $+Q,$ are parallel and separated by a distance $a .$ The first rod has one end at the origin and its other end on the positive $y$ -axis. The second rod has its lower end on the positive $x$ -axis. (a) Explain why the $y$ -component of the net force on the second rod vanishes. (b) Determine the $x$ -component of the differential force $d F_{2}$ exerted on a small portion of the second rod, with length $d y_{2}$ and position $y_{2},$ by the first rod. (This requires integrating over differential portions of the first rod, parameterized by $\left.d y_{1} .\right)$ (c) Determine the net force $\vec{F}_{2}$ on the second rod by integrating $d F_{2 x}$ over the second rod. (d) Show that in the limit $a \gg L$ the force determined in part (c) becomes $\frac{1}{4 \pi \epsilon_{0}} \frac{Q^{2}}{a^{2}} \hat{\imath}$. (e) Determine the external work required to move the second rod from very far away to the position $x=a$, provided the first rod is held fixed at $x=0 .$ This describes the potential energy of the original configuration.
(f) Suppose $L=50.0 \mathrm{~cm}, a=10.0 \mathrm{~cm}, Q=10.0 \mu \mathrm{C},$ and $m=500 \mathrm{~g}$. If the two rods are released from the original configuration, they will fly apart and ultimately achieve a particular relative speed. What is that relative speed?

Key Concepts

Symmetry and Component Cancellation

Symmetry in the arrangement of charged objects can lead to significant simplifications in the analysis of forces. For instance, symmetrical distributions can cause certain vector components, such as the vertical components of force in parallel rod arrangements, to cancel out. Recognizing symmetry helps in reducing the complexity of integration and identifying zero contributions without explicitly performing the complete calculation.

Coulomb's Law

Coulomb's Law provides the fundamental relationship between electric charges, stating that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. This concept is extended to distributed charges by considering the force between differential elements of charge and then integrating over the distribution.

Superposition Principle

The superposition principle in electrostatics asserts that the net force on any charge is the vector sum of the forces exerted by all other individual charges. This means that when dealing with continuous charge distributions or multiple charge elements, each contribution is calculated separately and then added, allowing complex charge interactions to be broken down into manageable parts.

Integration of Continuous Charge Distributions

When charges are spread out over a line, surface, or volume, the force and potential energy must be computed by integrating differential contributions from each infinitesimal element of charge. This method involves parameterizing the distribution, applying Coulomb's Law locally, and then summing the effects to obtain the net result, which is crucial in problems involving rods or other extended objects.

Electrostatic Potential Energy and Work

Electrostatic potential energy represents the work required to assemble a configuration of charges from infinity. Calculating this energy involves integrating the work done against the electrostatic forces as the charges are brought together. This concept is vital in understanding the energy changes that occur when a charge distribution is reconfigured, such as moving a rod from far away to a specific position.

Energy Conservation in Electrostatics

The principle of energy conservation in electrostatics governs the transformation of potential energy into kinetic energy and vice versa in isolated systems. In problems involving the dynamics of charged objects, the stored potential energy due to their configuration is converted into kinetic energy when the objects are released. This conversion is used to predict motions such as the relative speeds achieved by the objects as they fly apart after being released.

Transcript

00:01 So for part a, we have because both rods have uniform charge distribution, every charge element above the midpoint of rod 1 will have, and i have, and i have, and i, identical element below the midpoint.

01:01 For part b, we have df of 2x is equal to df of 2, cosine of theta, which is equal to 1 over 4 pi epsilon not, times dq1, dq2, times r12 squared, times cosine of theta, times cosine of theta, so, we know that df of 2x is equal to wavelength a over 4 pi epsilon knot times negative 1 over a squared times y2 minus y1 divided by the square root of y squared minus y2, or sorry, y2 minus y2, plus a squared times ly1 equals 0 dq of 2.

02:26 For c, we now integrate the result from part b.

02:30 So we have f of 2x equals negative 4 pi epsilon knot of a times a minus the square root of l squared plus a squared, minus the square root of l squared plus a squared minus a.

02:57 And that equals that over 2 pi, epsilon knot of a, times the square root of l squared plus a squared minus a.

03:16 So the force vector, then force vector, is equal to q squared over 2 pi.

03:27 Epsilon not l squared times the square root of l divided by a squared plus one minus one for part d we use the result from part c so f of 2x is equal to q squared over 2 pi epsilon not l squared times one plus one half divided by l divided by a squared minus one and that equals q squared over 4 pi epsilon knot of a squared.

04:15 So then for par e, we have w equals the limit of x as it approaches infinity of a and x, so f of x, dx.

04:34 And so we know w is equal to negative q squared over 2 pi abson knot of l squared times the integral of a and x times the square root of l squared plus x squared minus x over x times the derivative of x...

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