An infinitely long thin straight wire has uniform linear charge density of 1/3{cm}^{-1}. Then, the magnitude of the electric intensity at a point 18 cm away is (given ε₀=8.8×10^{-12}{C}²{Nm}^{-2})

Correct Answer is : 0.33×1011NC−10.33×10¹1}{NC}^{-1}0.33×1011NC−1

Correct Answer is : 0.33×1011NC−10.33×10¹1}{NC}^{-1}0.33×1011NC−1

Test Index

VITEEE 2009 Paper

Question : 26 of 120

Marks: +1, -0

\frac{1}{3}\mathrm{cm}^{-1}.

\varepsilon_{0}=8.8\times10^{-12}\mathrm{C}^{2}\mathrm{Nm}^{-2}

)

Solution:

Charge density or charge per unit length of long wire

\lambda=\frac{1}{3}\mathrm{Cm}^{-1}\text{ and }r=18\times10^{-2}\mathrm{m}

According to Gauss theorem,

\oint\overset{\rightarrow}{E}\cdot\overset{\rightarrow}{dS}=\frac{q}{\varepsilon_{0}}

⇒

E\oint dS=\frac{q}{\varepsilon_{0}}

E\times2\pi r l=\frac{q}{\varepsilon_{0}}

⇒

E=\frac{q}{2\pi\varepsilon_{0} r l}=\frac{q/l}{2\pi\varepsilon_{0} r}

=\frac{\lambda\times2}{2\pi\varepsilon_{0} r\times2}=\frac{\lambda\times2}{4\pi\varepsilon_{0} r}

=9\times10^{9}\times\frac{1}{3}\times2\times\frac{1}{18\times10^{-2}}

=0.33\times10^{11}\mathrm{NC}^{-1}

Go to Question:

Prev Question

Next Question