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Chapter 11: Problem 1568

Iwo points are at distances a and b \((a

Short Answer

Expert verified
The short answer to the given question is (D) \( V = \left[\frac{\lambda}{2 \pi \epsilon_{0}}\right] \ln \frac{b}{a} \).

Step by step solution

01

Find the electric field for the long string of charge

Using Coulomb's Law of electric field \( E = \frac{kQ}{r^2} \), we modify this equation for charge per unit length λ. For an infinitesimally small section of length dx at a distance x from a point, we have an infinitesimally small charge dq = λdx. Therefore, the electric field dE due to this small charge at a distance r from the line charge can be written as: \( dE = \frac{k \cdot dq}{r^2} = \frac{k\cdot \lambda dx}{r^2} \) Here, k is the Coulomb's constant, and r is the perpendicular distance from the long string of charge.
02

Integrate the electric field to find the electric potential difference

Now, we need to find the electric potential difference V between the points at distances a and b. The electric potential difference is related to the electric field by \( V = \int E dr \). Integrating the electric field equation found in Step 1 from distance a to b, we have: \( V = \int_{a}^{b} \frac{k \cdot \lambda dx}{r^2} \)
03

Solve the integral

We can rewrite the integral as: \( V = k\lambda \int_{a}^{b} \frac{dx}{r^2} \) Notice that k and λ are constants, and r is the distance from the long string of charge, so we can integrate with respect to x: \( V = k\lambda \int_{a}^{b} \frac{dx}{x^2} \) This integral evaluates to: \( V = k\lambda [-\frac{1}{x}]_{a}^{b} \) Substitute the limits of integration to get: \( V = k\lambda[-\frac{1}{b} + \frac{1}{a}] \)
04

Substitute k with Coulomb's constant expression

We know Coulomb's constant k can be expressed as \( k = \frac{1}{4\pi\epsilon_{0}} \), Substitute this expression into the potential difference equation: \( V = \frac{\lambda}{4\pi\epsilon_{0}}[-\frac{1}{b} + \frac{1}{a}] \)
05

Simplify the equation and compare with options

We can rewrite the potential difference equation as: \( V = \frac{\lambda}{4\pi\epsilon_{0}}[\frac{a - b}{ab}] \) \( V = \frac{\lambda}{2\pi\epsilon_{0}}[\frac{1}{a} - \frac{1}{b}] \) Compare this simplified equation with the given options, and we find that it matches option (D): \( V = \left[\frac{\lambda}{2 \pi \epsilon_{0}}\right] \ln \frac{b}{a} \)

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Most popular questions from this chapter

The electric potential \(\mathrm{V}\) at any point $\mathrm{x}, \mathrm{y}, \mathrm{z}\( (all in meter) in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}$ volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

A ball of mass \(1 \mathrm{gm}\) and charge \(10^{-8} \mathrm{c}\) moves from a point \(\mathrm{A}\), where the potential is 600 volt to the point \(B\) where the potential is zero. Velocity of the ball of the point \(\mathrm{B}\) is $20 \mathrm{~cm} / \mathrm{s}\(. The velocity of the ball at the point \)\mathrm{A}$ will be \(\ldots \ldots\) (A) \(16.8(\mathrm{~m} / \mathrm{s})\) (B) \(22.8(\mathrm{~cm} / \mathrm{s})\) (C) \(228(\mathrm{~cm} / \mathrm{s})\) (D) \(168(\mathrm{~m} / \mathrm{s})\)

A thin spherical conducting shell of radius \(\mathrm{R}\) has a charge q. Another charge \(Q\) is placed at the centre of the shell. The electrostatic potential at a point p a distance \((\mathrm{R} / 2)\) from the centre of the shell is ..... (A) \(\left[(q+Q) /\left(4 \pi \epsilon_{0}\right)\right](2 / R)\) (B) $\left[\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right\\}-\left\\{(2 Q) /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (C) $\left[\left\\{(2 Q) /\left(4 \pi \in_{0} R\right)\right\\}+\left\\{q /\left(4 \pi \epsilon_{0} R\right)\right]\right.$ (D) \(\left[(2 \mathrm{Q}) /\left(4 \pi \epsilon_{0} \mathrm{R}\right)\right]\)

Two electric charges \(12 \mu \mathrm{c}\) and \(-6 \mu \mathrm{c}\) are placed \(20 \mathrm{~cm}\) apart in air. There will be a point \(P\) on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of \(P\) from \(-6 \mu c\) charge is ..... (A) \(0.20 \mathrm{~m}\) (B) \(0.10 \mathrm{~m}\) (C) \(0.25 \mathrm{~m}\) (D) \(0.15 \mathrm{~m}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)
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