Question

The linear charge density of straight line
charge of infinite length is 1 uc/m. Find the
electric field intensity at a distance 1 m from
the line charge.
​

Answer 1

Given:

Linear charge density of straight line charge of infinite length (λ) = 10 μC/m = $\sf 10 \times 10^{-6} \ C/m$

Distance from the line charge (r) = 1 m

To Find:

Electric field intensity due to line charge

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:

$\rm \implies E = \dfrac{10 \times {10}^{ - 5} \times 2 \times 9 \times {10}^{9} }{ 1} \\ \\ \rm \implies E = 2 \times 9 \times {10}^{9 + 1 - 5} \\ \\ \rm \implies E = 18 \times {10}^{5} \: N/C \\ \\ \rm \implies E = 18 \times {10}^{6} \: N/C \\ \\ \rm \implies E = 1.8 \: MN/C$

$\therefore$ $\boxed{\mathfrak{Electric \ field \ intensity \ due \ to \ line \ charge \ (E) = 1.8 \ MN/C}}$

Answer 2

Answer:

E = 1.8 MN/C

Explanation:

Given that,

• The linear charge density of straight line charge of infinite length (λ) = 1uc/m → 10 × $</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>l</strong><strong>d</strong><strong>{</strong><strong>10^{-6}</strong><strong>}</strong><strong>$C/m

• Displacement (r) = 1m

As we know that,

Electric field at a distance r from infinitely long charged wire ;

${ \boxed{ \red{ \underline { \frak{E = \frac{λ}{2 \pi ϵ_0r} }}}}}$

[ Putting values ]

$\implies \sf \: E = \frac{10 \times 10 {}^{ - 5} }{2 \times 9 \times 10 {}^{9} \times 1}$

$\implies \sf \: E = \frac{10 \times 2 \times 9 \times 10 {}^{9 + 1 - 5} }{1}$

$\implies \sf \: E =18 \times 10 {}^{5}$

$\implies \sf \: \green{E =1.8 \:MN/C}$

Hence,

• The Electric field intensity due to line charge is 1.8 MN/C.