# Electric Field due to line charge

**1. The problem statement, all variables and given/known data**

A non-conducting rod AB, having uniformly distributed positive charge of linear charge density λ is kept in x-y plane. The rod AB is inclined at an angle 30° with +ve Y-axis. The magnitude of electrostatic field at origin due to rod AB is E_0 N/C and its direction is along line OC. If line OC makes an angle θ=10a+b degree with negative x-axis as shown in the figure, calculate the value of (a+b) [OA=2m and λ=10^3 C/m]

**2. Relevant equations**

Please see the attached diagram

**3. The attempt at a solution**

[itex]E_x = \dfrac{\lambda}{4 \pi \epsilon _0 d} (\sin 60 + \sin 30) \\

E_y = \dfrac{\lambda}{4 \pi \epsilon _0 d} |\cos 30 – \cos 60| \\

\tan \alpha = \dfrac{E_y}{E_x} \\

=\dfrac{ |\cos 30 – \cos 60| }{(\sin 60 + \sin 30)} [/itex]

which comes out to be 15°.

Thus, θ = 30° – 15° = 15°. So, a+b = 6. But it’s not the correct answer 🙁

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