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

Key concepts

Electric Field

The electric field is a fundamental concept in electrostatics. It describes the force per unit charge exerted at any point around a charge distribution. It's like the invisible "force field" produced by electrically charged objects. The strength of the electric field, denoted as \( E \), is defined by the equation:\[ E = \frac{F}{q} \]Here, \( F \) is the force experienced by a small positive test charge \( q \) at a given location in the field. For a point charge, the field vector points radially outward from a positive charge and inward toward a negative charge. In the context of a line of charge, as in the exercise, the field is not uniform and varies with distance. When dealing with a linear charge density \( \lambda \), the electric field created by this line charge is derived differently, as shown in the original solution. We consider small elements of the line, each producing their own small electric field, and integrate these contributions over the length of the line to obtain the total electric field.

Electric Potential

Electric potential is an essential electrostatic concept, representing the potential energy per unit charge at a certain position in an electric field. The electric potential difference between two points is a scalar quantity and is a measure of the work done in moving a unit positive charge from one point to another within the field.Mathematically, potential difference \( V \) can be expressed as the negative integral of the electric field:\[ V = -\int E \cdot dr \]This tells us that potential difference depends on the path taken between two points in space. However, in electrostatics where electric fields are conservative, it is path independent and only depends on the initial and final positions. In the case of a line charge, calculating the potential involves integrating the electric field from distance \( a \) to \( b \), leading to the result seen in the original exercise. The potential difference becomes logarithmic in form when dealing with linear charge distributions, reflecting the natural decrease in electric field intensity with distance.

Coulomb's Law

Coulomb's Law is the cornerstone of electrostatics. It describes how the electric force between two stationary point charges depends on their separation and magnitudes. The law can be formulated as:\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the magnitude of the force between two charges \( q_1 \) and \( q_2 \), \( r \) is the distance between them, and \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \)).In applications beyond point charges, such as with an infinite line charge, Coulomb’s Law helps determine how these charges contribute to the electric field at a point in space. In the exercise solution, Coulomb's Law is adapted for a continuous line of charge by considering a tiny piece of the charge and then integrating over the entire length to find the total electric effect.

• Coulomb's Law helps express how charges interact with one another.
• It is especially crucial in calculating forces and fields in more complex charge distributions.