Question

A wire AMB is placed along y-axis having linear
charge density , C/m. The electric field intensity at
point Pis (PM = 0.1 m)
B
377
37
P​

Answer 1

Answer:

This is your answer there are 4 photos

Answer 2

Answer:

The electric field intensity is 1.44×10¹¹λ(V/m).

Explanation:

Given:

Linear charge density = λ C/m

PM (R) = 0.1m

To find:

Electric field intensity at P

Solution:

We can see that on the outer surface of wire AMB, there are positive charges and line PM is perpendicular to AMB.

∠BPM = ∠APM = 37°

We know that the electric field intensity at a point is given as

E = (kλ/R)(cosθ1 - cosθ2)

Here, θ1 = 37° and θ2 = 180° - θ1

E = (kλ/R)[cos37° - cos(180° - 37°)]

E = (kλ/R)[cos37° - (-cos37°)]

E = (2kλ/R) (cos37°)

Substituting the values of k and R, we get

E = (2×9×10⁹×λ×0.799)/0.1

E = (1.44×10¹⁰×λ)/0.1

E = 1.44×10¹¹λ(V/m)

Therefore, the electric field intensity at point P is 1.44×10¹¹λ(V/m).

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