Question

In the figure the electric field due to a finite line of charge with linear charge density lambda=10^(-8)C/m at point P is (given d =50cm)
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Answer 1

In the figure, the electric field due to a finite line of charge with linear charge density, λ = 10^-8 C/m at point P (given d = 50 cm) is ...

solution : cut an element dx , x distance from the line of point P.

charge on element, dQ = λdx

so electric field due to element at point P, dE = $K\frac{dQ}{(\sqrt{x^2+d^2})^2}\frac{d}{\sqrt{x^2+d^2}}$

= $K\lambda\int\limits^{a}_{-a}\frac{dx}{(x^2+d^2)^{3/2}}$

= $\frac{K\lambda}{d}\left[\frac{x}{\sqrt{x^2+d^2}}\right]^a_{-a}$

= $\frac{k\lambda}{d}\left[\frac{2a}{\sqrt{a^2+d^2}}\right]$

here a = d/√3

= $\frac{k\lambda}{d}$

now putting d = 50 cm = 0.5 m, k = 9 × 10^9 Nm²/C² , λ = 10^-8 C/m

so, E = (9 × 10^9 × 10^-8)/(0.5) = 180 N/C

Therefore the electric field at point P is 180 N/C.

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