Question

Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2×10-6 C m-1.

Answer 1

Answer:

Explanation:

Electric field intensity in case of infinite long wire at a distance r is given by

E =

Here, λ is linear charge density .

given,

λ = 2 × 10⁻⁶ C/m and r = 4cm = 0.04 m

Hence , E =

= 2 × 10⁻⁶/(2 × 3.14 × 8.85 × 10⁻¹² × 0.04)

= 10⁸/(3.14 × 8.85 × 4) N/C

= 0.0899 × 10⁷ N/C

= 8.99 × 10⁵ N/C

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Answer 2

The magnitude of the electric field is $8.99 \times 10^{6} \mathrm{N} / \mathrm{C}$

Explanation:

Electric Field Intensity = $\lambda / 2 \pi \varepsilon_{0} r(\mathrm{Known})$

Provided ⇒

Density of linear charge  $(\lambda)=2 \times 10^{-6} \mathrm{c} / \mathrm{m}$

Distance(r) = 4 cm.

= 0.04 m.

Also, $\varepsilon_{0}=8.85 \times 10^{-12}$

We will proceed using the Formula,

Electric Field Intensity(E)  $=\lambda / 2 \pi \varepsilon_{0} r$

$\begin{array}{l}{\therefore \mathrm{E}=\left(2 \times 10^{-6}\right) \div\left(2 \times 22 / 7 \times 8.85 \times 10^{-12} \times 0.04\right)} \\{\Rightarrow \mathrm{E}=10^{6} / 1.1126} \\{\therefore \mathrm{E}=0.899 \times 10^{6}} \\{\therefore \mathrm{E}=8.99 \times 10^{6} \mathrm{N} / \mathrm{C}}\end{array}$

Hence, $8.99 \times 10^{6} \mathrm{N} / \mathrm{C}$ is the intensity of the Electric Field Intensity.