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A thin rod extends along the \(z\) axis from \(z=-d\) to \(z=d\). The rod carries a charge uniformly distributed along its length with linear charge density \(\lambda\). By integrating over this charge distribution calculate the potential at a point \(P_{1}\) on the \(z\) axis with coordinates 0 . \(0,2 d\). By another integration find the potential at a point \(P_{2}\) on the \(x\) axis and locate this point to make the potential equal to the potential at \(P_{1}\)

Short Answer

Expert verified
The potential at point P1 is equal to the potential at point P2 when P2 is at (√3d, 0, 0).

Step by step solution

01

Define the Problem

We need to find the electric potential at two points: one on the z axis at (0, 0, 2d) and another on the x-axis such that the potentials at both points are equal. The charge density is given as \( \lambda \), and the rod extends from \( z = -d \) to \( z = d \).
02

Expression for Potential

The potential at a point due to a small charge element \( dq \) is given by \( dV = \frac{k_e \, dq}{r} \), where \( k_e \) is Coulomb's constant and \( r \) is the distance from the charge element to the point. For a linear charge with linear density \( \lambda \), \( dq = \lambda \, dz \).
03

Calculate Potential at Point P1

At point \( P_1 (0, 0, 2d) \), the distance to a charge element at \( z \) is \( r = |2d - z| \). Substitute into the potential formula: \[ V_{1} = \int_{-d}^{d} \frac{k_e \lambda}{|2d - z|} \, dz \]. Solve this integral to find the potential.
04

Calculate and Integrate for V1

Integrate the expression from Step 3: \[ V_{1} = k_e \lambda \left( \ln(|2d - z|) \right)_{-d}^{d} \] This evaluates to: \[ V_{1} = k_e \lambda \left( \ln(d) - \ln(3d) \right) = -k_e \lambda \ln(3) \].
05

Setup for Potential at Point P2

For point \( P_2 (x, 0, 0) \), the distance to a charge element at \( z \) is \( r = \sqrt{x^2 + z^2} \). The potential is \[ V_{2} = \int_{-d}^{d} \frac{k_e \lambda}{\sqrt{x^2 + z^2}} \, dz \].
06

Calculate and Integrate for V2

Solve the integral for \( V_{2} \): \[ V_{2} = k_e \lambda \int_{-d}^{d} \frac{1}{\sqrt{x^2 + z^2}} \, dz \].Change variables by substituting \( z = x \tan(\theta) \) leading to \[ V_{2} = k_e \lambda \ln(\frac{\sqrt{d^2 + x^2} + d}{\sqrt{d^2 + x^2} - d}) = -k_e \lambda \ln(3) \].
07

Equal Potentials Solution

Equate \( V_{1} \) and \( V_{2} \) to find the necessary \( x \):\[ \ln(\frac{\sqrt{d^2 + x^2} + d}{\sqrt{d^2 + x^2} - d}) = \ln(3) \].Solving gives \( \frac{\sqrt{d^2 + x^2} + d}{\sqrt{d^2 + x^2} - d} = 3 \).
08

Solve for x

Solving the equation from Step 7: \( 3(\sqrt{d^2 + x^2} - d) = (\sqrt{d^2 + x^2} + d) \) leads to \( \sqrt{d^2 + x^2} = 2d \). Solving further gives \( x^2 + d^2 = 4d^2 \), hence \( x = \sqrt{3}d \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Charge Density
Linear charge density is a crucial concept when dealing with problems involving charged objects such as rods or wires. It is defined as the amount of electric charge per unit length along an object. If a rod has a uniform charge distribution, the linear charge density, denoted by \( \lambda \), is constant across the rod's entire length. This means that no matter which section of the rod you look at, the charge per unit length is the same.

For the thin rod extending from \( z = -d \) to \( z = d \), the linear charge density \( \lambda \) represents the uniform distribution of the rod's total charge \( Q \) over its length of \( 2d \). Therefore, \( \lambda \) can be expressed as \( \lambda = \frac{Q}{2d} \). This density is used in calculations involving the electric potential by considering small charge elements \( dq \), which is equal to \( \lambda \ dz \), where \( dz \) represents an infinitesimally small segment of the rod.
Understanding linear charge density helps in setting up the mathematical expressions that describe how the electric potential varies along different positions relative to the rod.
Integral Calculus
Integral calculus allows us to calculate quantities such as electric potential that arise from continuously distributed charges like those along a rod. When calculating the potential at a specific point due to a distributed charge, we must integrate because the charge is spread out over a length, area, or volume.

In this exercise, the potential at a point due to a continuous charge distribution along a rod is determined by integrating the contributions of infinitesimally small charge elements, \( dq = \lambda \ dz \), across the length of the rod. The formula used for one such element is \( dV = \frac{k_e \ dq}{r} \), where \( r \) is the distance from the charge element to the point of interest.
Integration serves to sum up all these small potentials into a single value for the entire rod. For example, the potential at point \( P_1 \) on the z-axis involves solving the integral \[ V_{1} = \int_{-d}^{d} \frac{k_e \lambda}{|2d - z|} \, dz \]. This process encapsulates the effect of the entire charge distribution in one calculation.
Coulomb's Law
Coulomb's Law is foundational in calculating electric fields and potentials. It states that the electric force \( F \) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The law is mathematically expressed as \( F = k_e \frac{q_1 q_2}{r^2} \), where \( k_e \) is Coulomb's constant.

When dealing with continuous charge distributions, such as a charged rod, we extend the principles of Coulomb's Law to find potentials by considering small charge segments \( dq \). The potential contribution from a small charge \( dq \) to a point at distance \( r \) is given by \( dV = \frac{k_e \ dq}{r} \).
This exercise highlights how Coulomb's Law is applied through integration. Each small element \( dq \) along the rod behaves like a point charge, and the potential from each is calculated and summed to find the total potential at a point using the principle of superposition. The results demonstrate how distributed charges can create complex fields and potentials, which are effectively handled using the tools provided by Coulomb's Law and calculus.

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Most popular questions from this chapter

Designate the corners of a square, \(5 \mathrm{~cm}\) on a side, in clockwise order, \(A, B, C, D .\) Put a charge 2 esu at \(A,-3\) esu at \(B .\) Determine the value of the line integral of \(\mathbf{E}\), from point \(C\) to point \(D\). (No actual integration needed!)

A sphere the size of the earth has 1 coulomb of charge distributed evenly over its surface. What is the electric field strength just outside the surface, in volts/meter? What is the potential of the sphere, in volts, with zero potential at infinity?

An interstellar dust grain, roughly spherical with a radius of \(3 \times 10^{-7}\) meters, has acquired a negative charge such that its potential is \(-0.15\) volt. How many extra electrons has it picked up? What is the strength of the electric field at its surface, expressed in volts/ meter?

By means of a van de Graaff generator, protons are accelerated through a potential difference of \(5 \times 10^{6}\) volts. The proton beam then passes through a thin silver foil. The atomic number of silver is 47, and you may assume that a silver nucleus is so massive compared with the proton that its motion may be neglected. What is the closest possible distance of approach, of any proton, to a silver nucleus? What will be the strength of the electric field acting on the proton at that position?

The vector function which follows represents a possible electrostatic field: $$ E_{x}=6 x y \quad E_{y}=3 x^{2}-3 y^{2} \quad E_{z}=0 $$ Calculate the line integral of \(\mathbf{E}\) from the point \((0,0,0)\) to the point \(\left(x_{1}, y_{1}, 0\right)\) along the path which runs straight from \((0,0,0)\) to \(\left(x_{1}, 0,\right.\), 0 ) and thence to \(\left(x_{1}, y_{1}, 0\right)\). Make a similar calculation for the path which runs along the other two sides of the rectangle, via the point \((0,\), \(y_{1}, 0\) ). You ought to get the same answer if the assertion above is true. Now you have the potential function \(\phi(x, y, z)\). Take the gradient of this function and see that you get back the components of the given field.

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