Question

An infinite line charge produces a field of 9 x 10-4 NC-1 at a distance of 2 cm. Calculate the
charge density
1 Point
sono
- 10 cm​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Charge\:Density=1\times 10^{-15}\:C\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Electric \: field = 9 \times {10}^{ - 4} \: NC^{ - 1} \\ \\ \tt: \implies Distance(r) = 2 \: cm \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Charge \: Density( \lambda) = ?$

• According to given question :

$\tt \circ \: r = \frac{2}{100} = 0.02 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies E = \frac{2k \lambda}{r} \\ \\ \tt \circ \: k = 9 \times {10}^{9} \\ \\ \tt: \implies 9 \times {10}^{ - 4} = \frac{2 \times 9 \times 10^{9} \times \lambda }{0.02 } \\ \\ \tt: \implies \frac{9 \times {10}^{ - 4} \times 0.02 }{2 \times 9 \times 10^{9} } = \lambda \\ \\ \tt: \implies \lambda = \frac{2 \times {10}^{ - 4 - 2 - 9} }{2} \\ \\ \green{\tt: \implies \lambda =1 \times {10}^{ - 15} \: C \: {m}}$

Answer 2

Given ,

Electric filed (E) = 9 × 10^(-4) N/C

Distance (r) = 2 cm or 2 × 10^(-2) m

We know that , the electric field due to an infinitely long charged wire is given by

$\boxed{ \sf{E = \frac{ \lambda}{2\pi e_{o}r } = \frac{2 K\lambda}{r} }}$

Where ,

K = 9 × 10^(9) Nm²/C²

Thus ,

$\sf \mapsto 9 \times {(10)}^{ - 4} = \frac{2 \times 9 \times {(10)}^{9} \times \lambda}{2 \times {(10)}^{ - 2} } \\ \\ \sf \mapsto \lambda = 1 \times {(10)}^{ - 15} \: \: C/m$

∴ The linear charge density is 10^(-15) C/m