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 ?
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 ?
Updated On: Jul 1, 2024
- $1.12 \times 10^8 \,NC^{-1}$
- $4.5 \times 10^8 \,NC^{-1}$
- $2.25 \times 10^8 \,NC^{-1}$
- $9 \times 10^8 \,NC^{-1}$
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The Correct Option is C
Solution and Explanation
We know that
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$
$E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
$=\frac{1}{4} \times \frac{10^{-2}}{10^{-2}} \times 18 \times 10^{9} \times 5$
$=2.25 \times 10^{8} N / C$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).