UNNECESSARY ANSWERS WILL BE REPORTED
Answers
Answer:
Electric field intensity due to infinite line charge (E) = \sf 9 \times 10^4 \ NC^{-1}9×10
4
NC
−1
Distance from the line charge (r) = 2 cm = 0.02
To Find:
Linear charge density \sf (\lambda)(λ)
Answer:
Electric field intensity due to a infinite long straight line charge is given as:
\boxed {\bf {E = \dfrac{\lambda}{2\pi \epsilon_0 r}}}
E=
2πϵ
0
r
λ
By substituting values we get:
\begin{gathered}\rm \implies \cancel{9} \times {10}^{4} = \dfrac{ \cancel{2} \times \cancel{9} \times {10}^{9} \times \lambda}{ \cancel{2} \times {10}^{ - 2} } \\ \\ \rm \implies {10}^{4} = \frac{ {10}^{9} \times \lambda}{ {10}^{ - 2} } \\ \\ \rm \implies \lambda = {10}^{4 - 2 - 9} \\ \\ \rm \implies \lambda = {10}^{ - 7} \: C {m}^{ - 1}\end{gathered}
⟹
9
×10
4
=
2
×10
−2
2
×
9
×10
9
×λ
⟹10
4
=
10
−2
10
9
×λ
⟹λ=10
4−2−9
⟹λ=10
−7
Cm
−1
\therefore∴ \boxed{\mathfrak{Linear \ charge \ density \ (\lambda) = 10^{-7} \ C/m}}
Linear charge density (λ)=10
−7
C/m
Answer:
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