Question

An infinitely long thin straight wire has uniform charge density of $\frac {1}{4} \times 10^{-2} cm^{-1}.$ What is the magnitude of electric field at a distance $20\, cm$ from the axis of the wire ?

• A. $1.12 \times 10^8 \,NC^{-1}$
• B. $4.5 \times 10^8 \,NC^{-1}$
• C. $2.25 \times 10^8 \,NC^{-1}$
• D. $9 \times 10^8 \,NC^{-1}$

Answer 1

The Correct Option is C

For an infinitely long wire with uniform linear charge density \(\lambda\), the electric field at a distance \(r\) from the wire is given by the formula: \[ E = \frac{2k_e \lambda}{r} \] Where: - \(k_e = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\) is Coulomb's constant, - \(\lambda = \frac{1}{4} \times 10^{-2} \, \text{C/m}\) is the charge density, - \(r = 0.2 \, \text{m}\) is the distance from the wire. Substituting the values into the formula: \[ E = \frac{2 \times 9 \times 10^9 \times \left( \frac{1}{4} \times 10^{-2} \right)}{0.2} \] \[ E = \frac{2 \times 9 \times 10^9 \times 0.0025}{0.2} = 2.25 \times 10^8 \, \text{NC}^{-1} \] Thus, the magnitude of the electric field at a distance 20 cm from the wire is \(2.25 \times 10^8 \, \text{NC}^{-1}\).

Answer 2

We know that
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$

Answer 3

The electric field due to an infinitely long straight wire with linear charge density \( \lambda \) at a distance \( r \) from the wire is given by the formula: \[ E = \frac{2 k_e \lambda}{r} \] where: - \( k_e = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \) is the Coulomb constant, - \( \lambda \) is the linear charge density, - \( r \) is the distance from the wire. Given: - \( \lambda = \frac{1}{4} \times 10^{-2} \, \text{cm}^{-1} = \frac{1}{4} \times 10^{-4} \, \text{m}^{-1} \), - \( r = 20 \, \text{cm} = 0.2 \, \text{m} \). Substitute the values into the equation: \[ E = \frac{2 \times 9 \times 10^9 \times \frac{1}{4} \times 10^{-4}}{0.2} \] \[ E = \frac{9 \times 10^9 \times 10^{-4}}{0.1} = \frac{9 \times 10^5}{0.1} = 2.25 \times 10^8 \, \text{N/C}. \] Thus, the magnitude of the electric field is \({2.25 \times 10^8 \, \text{NC}^{-1}} \).