Finding Electric Field due to Sheet of Charge

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SUMMARY

The discussion focuses on calculating the resultant electric field due to a charged sheet and a point charge. The sheet of charge has a surface charge density of ρs = -25 x 10^-25 C/m², while the point charge Q = 12.5 x 10^-6 C is located at (-5, 2, 3). The relevant equations used include Gauss' law for a point charge, E = (Q/4πε0R²)*R/R, and for a sheet of charge, E = (ρs/2ε0)*aR. The user is seeking assistance in determining the normal vector from the plane equation 6x + 4y + 3z = 12 to the point (-5, 10, 35).

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Homework Statement



A sheet of charge, ρs = -25 x 10^-25 C/m^2, intersects the x-axis @ x = 2, the y-axis @ y = 3, and the z-axis @ z = 4. Additionally a point charge of Q = 12.5 x 10^-6 C is located @ (-5, 2, 3). Find the resultant electric field due to the sheet and the point charge @ (-5, 10, 35).

Homework Equations



E = (Q/4πε0R^2)*R/R
E = (ρs/2ε0)*aR

Sorry if they are difficult to read, first equation is generic Gauss' law for a point charge in unit vector notation and the second equation is Gauss' law for a sheet of charge in relation to a unit normal vector.

The Attempt at a Solution



I found the contribution of the point charge using the first equation (25.04j + 100.18k) and started on the plane charge. My guess would be to find the vector that passes through the point (-5, 10, 35), convert it to a unit vector and then multiply by (ρs/2ε0), but I don't know how to find the normal vector from the point and the plane equation. If there's another solution, please let me know. I've computed the plane equation to be: 6x + 4y + 3z= 12.

Any help is much appreciated!
 
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You have the equation of the plane. What you need from that is the normal vector of the plane, which lies along the shortest vector from the plane to any point not in the plane. Do the coefficients in the equation of the plane tell you something?
 

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