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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

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

Two point charges repel each other with a force of \(100 \mathrm{~N}\). One of the charges is increased by \(10 \%\) and other is reduced by \(10 \%\). The new force of repulsion at the same distance would be \(\ldots \ldots \mathrm{N}\). (A) 121 (B) 100 (C) 99 (D) 89

Two point charges \(100 \mu \mathrm{c}\) and \(5 \mu \mathrm{c}\) are placed at points \(\mathrm{A}\) and \(B\) respectively with \(A B=40 \mathrm{~cm}\). The work done by external force in displacing the charge \(5 \mu \mathrm{c}\) from \(\mathrm{B}\) to \(\mathrm{C}\) where \(\mathrm{BC}=30 \mathrm{~cm}\), angle \(\mathrm{ABC}=(\pi / 2)\) and \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right]\) \(=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{c}^{2}\). (A) \(9 \mathrm{~J}\) (B) \((9 / 25) \mathrm{J}\) (C) \((81 / 20) \mathrm{J}\) (D) \(-(9 / 4) \mathrm{J}\)

In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. \(2400 \mathrm{v}\) between the plates. To keep a drop of half the radius stationary the potential difference had to be made $600 \mathrm{v}$. What is the charge on the second drop? (A) \([(3 Q) / 2]\) (B) \((\mathrm{Q} / 4)\) (C) \(Q\) (D) \((\mathrm{Q} / 2)\)
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