Surface Charge Density for Infinite Sheet of Charge

[eestep]

Homework Statement

An infinite sheet of charge with a surface charge density of +7.8 micro-Coulomb/m² lies on x-axis (technically in x-z plane). Another infinite sheet of charge with an unknown charge density lies on y-axis (technically in y-z plane). A positive point charge is placed in first quadrant and total electric force on this charge due to infinite sheets is directed at an 143 degree angle when measured counter-clockwise from positive x axis. What is surface charge density of sheet on y-axis in micro-Coulomb/m²? Answer is -10.35.

Homework Equations

\sigma=dq/dA

The Attempt at a Solution

F_ex=-qE=-q\sigma_x/2\epsilon₀
F_ey=qE=q\sigma_y/2\epsilon₀
tan\theta=(q\sigma_y/2\epsilon₀)/(q\sigma_x/2\epsilon₀)
tan\theta=\sigma_y/\sigma_x
\sigma_y=\sigma_xtan\theta=7.8tan143

 

[eestep]

Does it not cancel anyhow?

 

[SammyS]

Homework Statement

An infinite sheet of charge with a surface charge density of +7.8 micro-Coulomb/m² lies on x-axis (technically in x-z plane). Another infinite sheet of charge with an unknown charge density lies on y-axis (technically in y-z plane). A positive point charge is placed in first quadrant and total electric force on this charge due to infinite sheets is directed at an 143 degree angle when measured counter-clockwise from positive x axis. What is surface charge density of sheet on y-axis in micro-Coulomb/m²? Answer is -10.35.

Homework Equations

\sigma=dq/dA

The Attempt at a Solution

F_ex=-qE=-q\sigma_x/2\epsilon₀ This should be F_ey .
F_ey=qE=q\sigma_y/2\epsilon₀ This should be F_ex .
tan\theta=(q\sigma_y/2\epsilon₀)/(q\sigma_x/2\epsilon₀)
tan\theta=\sigma_y/\sigma_x
\sigma_y=\sigma_xtan\theta=7.8tan143

The sheet with known charge, lying in the x-z plane, produces an electric field parallel to the y-axis.

 

[eestep]

How do I calculate surface charge density of sheet on y axis?

 

[SammyS]

How do I calculate surface charge density of sheet on y axis?

The vector you said was in the x direction is in the y direction.

The vector you said was in the y direction is in the x direction.

Correcting this will give you the right answer.

 

[eestep]

I appreciate your assistance!