Is the Electrostatic Force Calculation Correct for Inclined Bar and Wire?

[hastings]

find electrostatic force

problem:
there's a long straight wire with a λ1 charge. at some distance from it there's a bar charged with a λ2 charge. the bar is not parallel to the straight wire, instead it's inclined with an angle "alpha" between the dotted line, parallel to the wire, passing through the middle of the bar(point A), and the bar itself. The entire bar is long L. the distance between the wire and point A, is H.

Please check out the diagram.

What I did:

$$E=\frac{\lambda_1}{2\pi\epsilon_0 x}$$ this is the electrical field generated by the wire.

$$dq_2=\lambda_2 dl$$ . is this right? Then...

$$dx=dl \sin\alpha \longrightarrow dl=\frac{1}{\sin\alpha}dx$$ . if this is right then I substitute in the above relation:

$$dq_2=\lambda_2 dl=\lambda_2\frac{1}{\sin\alpha}dx$$ Is this correct till this point?

$$F=\int_{H- \frac{L}{2}\sin\alpha}^{H+\frac{L}{2}\sin\alpha} dF$$

Are the extremes of integration right?

Now the big problem is putting the pieces together.

$$F=\int_{H-\frac{L}{2}\sin\alpha}^{H+\frac{L}{2}\sin\alpha}dF= \int_{H-\frac{L}{2}\sin\alpha}^{H+\frac{L}{2}\sin\alpha}Edq_2=\int_{H-\frac{L}{2}\sin\alpha}^{H+\frac{L}{2}\sin\alpha}\frac{\lambda_1}{2\pi\epsilon_0 x}\lambda_2dl=\int_{H-\frac{L}{2}\sin\alpha}^{H+\frac{L}{2}\sin\alpha}\frac{\lambda_1}{2\pi\epsilon_0 x}(\lambda_2 \frac{1}{\sin\alpha }dx)$$


Could you tell me whether I need to fix anything?

 

[hastings]

Is it correct? Is it wrong?