# Finding the electric flux through the right face, confused on integration!

by mr_coffee
Tags: confused, electric, face, flux, integration
 Mentor P: 41,555 The only component of the field that contributes to the flux through a side is the component perpendicular to that side. For the right side of the cube, that perpendicular direction is the $\hat i$ direction. The component of the field in that direction is $3.0 x \hat i$; at x = 3 m, that component equals $9.0 \hat i$ (in units of N/C). Since the field is constant over the area of the right side, no integration is needed, just flux = E times Area.
 P: 1,629 ohhh i think i finally get it... so because the y component of the electric feild doesn't matter (4.0j), you can just discard it and only worry about the 3.0xi. and because x = 3, you end up with 9.0i. So really is i just telling the direction of the vector? you can just discard it? I'm still confused on one issue though. $\zeta [(3.0x)(dA)\hat i \bullet \hat i]$ You said you took the dot product, if A is pointing to the right, and also the electric feild is point right, wouldn't that be cos(0) = 1? how did they get 0? $\zeta [(3.0x)(dA) + 0]$ Sorry i'm really really rusty on vectors! that zeta is suppose to be an integral sign, i can't find the integral on the latex guide.
 HW Helper P: 2,277 It looks like you don't know this: $$\vec{i} \cdot \vec{i} = \vec{j} \cdot \vec{j} = \vec{k} \cdot \vec{k} = 1$$ $$\vec{i} \cdot \vec{j} = \vec{j} \cdot \vec{k} = \vec{i} \cdot \vec{k} = 0$$ Ah and the integral is $$\int$$