Question

an infinte line charged produces a field of 9*10power4 N/C at a distance of 2 cm . calculate the linear charge density

Answer 1

Given :

→ Field intensity = 9×10$^4$N/C

→ Distance = 2cm

To Find :

→ Linear charge density of infinite line charged conductor.

Formula :

☞ Formula of field intensity at a distance of r from a infinite line charged conductor is given by

$\bigstar\:\underline{\boxed{\bf{\red{\vec{E}=\dfrac{2k\lambda}{r}\hat{r}}}}}$

☆ Note :

• $\lambda$ is linear charge density (assumed uniform).
• r is perpendicular distance of point from line charge.
• $\hat{r}$ is radical unit vector drawn from the charge to test point.

Conversion :

▪ 100cm = 1m

▪ 2cm = 0.02m

Calculation :

$\Rightarrow\sf\:E=\dfrac{2k\lambda}{r}\\ \\ \Rightarrow\sf\:9\times 10^4=\dfrac{2\times (9\times 10^9)\times \lambda}{0.02}\\ \\ \Rightarrow\sf\:\lambda=\dfrac{9\times 10^4\times 0.02}{2\times 9\times 10^9}\\ \\ \Rightarrow\underline{\boxed{\bf{\gray{\lambda=1\times 10^{-7}\:Cm^{-1}}}}}$

Answer 2

Explanation:

$\bf \underline{Question}$

an infinte line charged produces a field of 9*10power4 N/C at a distance of 2 cm . calculate the linear charge density

_____________________________

$\bf \underline{Given}$

• line charged produces a field of 9×10
• power 4 N/C
• at a distance of 2 cm .

_____________________________

$\bf \underline{To\:Find}$

• the linear charge density

_____________________________

$\bf \underline{Solve}$

Now consider the linear charge density

$\sf\green{\frac{\lambda}{m}}$

linear charge density at a distance d is given by question.

$\sf\blue{t=\frac{\lambda}{2 \pi \epsilon_{o} d}…………eqn1}$

Here λ is the linear charge density,  

$\sf\epsilon_{o}$is the permittivity in the free space=

$\sf\blue{8.85 \times 10^{-12} n^{-1} m^{-2} c^{2}}$

Accourding to the question:

d is the distance

We solve According to the equation the)

$\sf\green{\lambda=t \times 2 \times \pi \in_{o} d}$

Putting all values

$\sf\blue{\lambda=9 \times 10^{4} \times 2 \times 3.14 \times 8.85 \times 10^{-12}}$

By solving:-

simplified:-

$\sf\red{\lambda=10 \mu c m^{-1}}$

Your answer is$\sf\pink{\lambda=10 \mu c m^{-1}}$