mr_coffee
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Hello again. The first problem I even came close to getting right, and yet i still missed it by a negative sign. There is a square surface measures 3.2mm on each side. It is immersed in a uniform elelctric field with magnitude E = 1800 N/C and with field lines at an angle of 35 degrees with normal to surface, as shown. Take that normal to be directed "outward" as though the surface were one face of a box. Calculate the electric flux through the surface.
So I used Flux = E cos(35)A; Plugged and chugged and got Flux = .015 Nm^2/C. The answer in the back of the book is Flux = -.015 Nm^2/C. Why would it be negative? If u need the image tell me and i'll scan it. It looks like a top of a box. Normal is straight up from the boxes surface pointing to the sky. the elctrical field lines are going through the box South Westward, you know if u draw a heart and put an arrow through it, that's what the lines look like they are doing.
So I used Flux = E cos(35)A; Plugged and chugged and got Flux = .015 Nm^2/C. The answer in the back of the book is Flux = -.015 Nm^2/C. Why would it be negative? If u need the image tell me and i'll scan it. It looks like a top of a box. Normal is straight up from the boxes surface pointing to the sky. the elctrical field lines are going through the box South Westward, you know if u draw a heart and put an arrow through it, that's what the lines look like they are doing.