Question

SPAMMING WILL BE REPORTED WITHOUT ANY DELAY​

Answer 1

Given:

Electric field intensity due to infinite line charge (E) = $\sf 9 \times 10^4 \ NC^{-1}$

Distance from the line charge (r) = 2 cm = 0.02

To Find:

Linear charge density $\sf (\lambda)$

Answer:

Electric field intensity due to a infinite long straight line charge is given as:

$\boxed {\bf {E = \dfrac{\lambda}{2\pi \epsilon_0 r}}}$

By substituting values we get:

$\rm \implies \cancel{9} \times {10}^{4} = \dfrac{ \cancel{2} \times \cancel{9} \times {10}^{9} \times \lambda}{ \cancel{2} \times {10}^{ - 2} } \\ \\ \rm \implies {10}^{4} = \frac{ {10}^{9} \times \lambda}{ {10}^{ - 2} } \\ \\ \rm \implies \lambda = {10}^{4 - 2 - 9} \\ \\ \rm \implies \lambda = {10}^{ - 7} \: C {m}^{ - 1}$

$\therefore$ $\boxed{\mathfrak{Linear \ charge \ density \ (\lambda) = 10^{-7} \ C/m}}$

Answer 2

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