Question

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

Short answer

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

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.

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: \( V = \int_{a}^{b} \frac{k \cdot \lambda dx}{r^2} \)

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: \( V = k\lambda[-\frac{1}{b} + \frac{1}{a}] \)

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: \( V = \frac{\lambda}{4\pi\epsilon_{0}}[-\frac{1}{b} + \frac{1}{a}] \)

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): \( V = \left[\frac{\lambda}{2 \pi \epsilon_{0}}\right] \ln \frac{b}{a} \)