Question

An infinitely long straight wire has uniform linear charge density of \(\dfrac{1}{3}\,cm^{-1}\). Then, the magnitude of the electric intensity at a point \(18\,cm\) away is (given \(\varepsilon_0 = 8.8 \times 10^{-12}\,C^2N^{-1}m^{-2}\))

• A. \(0.33 \times 10^{11}\,N\,C^{-1}\)
• B. \(3 \times 10^{11}\,N\,C^{-1}\)
• C. \(0.66 \times 10^{11}\,N\,C^{-1}\)
• D. \(1.32 \times 10^{11}\,N\,C^{-1}\)

Hint:

Electric field due to infinite line charge: \(E=\frac{\lambda}{2\pi\varepsilon_0 r}\). Always convert cm to m carefully.

Answer 1

The Correct Option is A

Step 1: Electric field due to infinite line charge.
\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]
Step 2: Convert \(\lambda\) into SI.
Given: \(\lambda = \frac{1}{3}\,C\,cm^{-1}\).
Convert to \(C\,m^{-1}\):
\[ \lambda = \frac{1}{3}\times 100 = \frac{100}{3}\,C\,m^{-1} \]
Step 3: Convert \(r\) into meters.
\[ r = 18\,cm = 0.18\,m \]
Step 4: Substitute values.
\[ E = \frac{\frac{100}{3}}{2\pi(8.8\times 10^{-12})(0.18)} \]
\[ E \approx \frac{33.33}{(2\pi)(1.584\times 10^{-12})} \]
\[ E \approx \frac{33.33}{9.95\times 10^{-12}} \approx 3.35\times 10^{12} \]
\[ E \approx 0.33\times 10^{13} = 0.33\times 10^{11}\,N\,C^{-1} \]
Final Answer: \[ \boxed{0.33 \times 10^{11}\,N\,C^{-1}} \]