Question

An infinite line of charge with charge density λ1 = 0.6 μC/cm is aligned with the y-axis.
What is Ex(P), the value of the x-component of the electric field produced by the line of charge at point P which is located at (x,y) = (a,0), where a = 7.5 cm?

Answer 1

Answer:

1440 × 10⁴ N/C

Explanation:

Data provided in the question:

Charge density λ1 = 0.6 μC/cm = 6 × 10⁻⁵ C/m

a = 7.5 cm = 0.075 m

Now,

Electric field due to a line charge at a distance 'a' is given as:

Ex(P) = $\frac{\lambda}{2\pi\epsilon_0a}$

also,

we know

$\frac{1}{4\pi\epsilon_0}$ = 9 × 10⁹ Nm²/C²

Thus,

we have

Ex(P) = $\frac{2}{2}\times\frac{\lambda}{2\pi\epsilon_0a}=\frac{2\lambda}{4\pi\epsilon_0a}$

therefore,

Ex(P) = $\frac{2\times6\times10^{-5}\times9\times10^9}{0.075}$

= 1440 × 10⁴ N/C

Answer 2

The value of the x-component of the electric field produced by the line of charge at point P is  $1440 \times 10^{4} \;\rm N/C$.

Charge Density is termed as the amount of electric charge present in any conductor per unit of length.

Given data:

The value of charge density is,  λ1 = 0.6 μC/cm.

The location of charge at a point P is, (x, y) = (a, 0).

Here, the value of a is 7.5 cm.

The expression for the electric field due to a line charge at a distance 'a' is given as,

Here, k is Coulomb's constant.

Solving as,

$E = \dfrac{2 \times (6 \times 10^{-5}) \times (9 \times 10^{9})}{0.075}\\\\ E = 1440 \times 10^{4} \;\rm N/C$

Thus, we can conclude that the value of the x-component of the electric field produced by the line of charge at point P is  $1440 \times 10^{4} \;\rm N/C$.

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