Question

The electric field due to an infinitely long thin straight wire with uniform linear charge density of \(2.5 \times 10^{-7}\) Cm\(^{-1}\) at a radial distance of x from the wire is \(7.5 \times 10^4\) NC\(^{-1}\). Then x =

• A. 2 cm
• B. 3 cm
• C. 4 cm
• D. 6 cm

Hint:

The electric field formula for a line charge, \(E = \frac{2k\lambda}{r}\), is essential. Using \(k = 9 \times 10^9\) is often much simpler than using \(\epsilon_0\), as it avoids dealing with \(\pi\). Always ensure your units are consistent (SI units are standard) before calculation.

Answer 1

The Correct Option is D

The formula for the electric field (E) at a radial distance (x) from an infinitely long straight wire with uniform linear charge density (\(\lambda\)) is given by Gauss's law:
\( E = \frac{\lambda}{2\pi\epsilon_0 x} \).
This formula can be more conveniently written using Coulomb's constant, \( k = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \) Nm\(^2\)C\(^{-2}\).
To do this, multiply the numerator and denominator by 2: \( E = \frac{2\lambda}{4\pi\epsilon_0 x} = \frac{2k\lambda}{x} \).
We are given the following values:
Electric field, \( E = 7.5 \times 10^4 \) NC\(^{-1}\).
Linear charge density, \( \lambda = 2.5 \times 10^{-7} \) Cm\(^{-1}\).
Coulomb's constant, \( k = 9 \times 10^9 \) Nm\(^2\)C\(^{-2}\).
We need to find the distance x. Rearrange the formula to solve for x:
\( x = \frac{2k\lambda}{E} \).
Substitute the given values into this equation.
\( x = \frac{2 \times (9 \times 10^9) \times (2.5 \times 10^{-7})}{7.5 \times 10^4} \).
\( x = \frac{18 \times 2.5 \times 10^{9-7}}{7.5 \times 10^4} = \frac{45 \times 10^2}{7.5 \times 10^4} \).
\( x = \frac{45}{7.5} \times 10^{2-4} = 6 \times 10^{-2} \) m.
The question asks for the answer in cm.
\( x = 6 \times 10^{-2} \text{ m} = 6 \text{ cm} \).