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Fall 2011 Homework 5 Solutions

# Fall 2011 Homework 5 Solutions - Physics 21 Fall 2011...

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Physics 21 Fall, 2011 Solution to HW-5 22-4 A cube has sides of length L = 0 . 330 m . It is placed with one corner at the origin as shown in the figure. The electric field is not uniform but is given by E = [ 4 . 41N / (Cm)] x ˆ i + [3 . 29N / (Cm] z ˆ k . (a) Find the electric flux through each of the six cube faces S 1 ,S 2 ,S 3 ,S 4 ,S 5 ,S 6 . (b) Find the total electric charge inside the cube. (a) The electric flux is defined as Φ = integraltext E · d A . Because d A is always perpendicular to the surface, we only need the component of the field that is normal to the surface of in- terest. (The dot product of d A with the parallel component of the field will be zero.) For each of the six surfaces, it will work out that the component normal to the surface is con- stant, so the surface integral will just be the constant value of the normal component times the surface area L 2 . For surface S 1 , there is no component of the electric field normal to the surface (in the y direction ˆ j ), so the electric flux Φ 1 = 0. For surface S 2 , the component of the electric field normal to the surface is in the + z direction, so take only the ˆ k component of E . The integral over the surface is integraldisplay L 0 integraldisplay L 0 3 . 29 z dxdy = 3 . 29 zL 2 . To evaluate the electric field at the surface we set z = L , and then the flux is Φ 2 = 3 . 29 L 3 = 0 . 118Nm 2 / C Surface S 3 is similar to S 1 ; there is no electric field compo- nent in the normal direction ˆ j , so the flux Φ 3 = 0. For surface S 4 , the normal direction is in the z direction, so we need to take the negative of the ˆ k component. The flux is that component integrated over the surface. Φ 4 = integraldisplay L 0 integraldisplay L 0 3 . 29 z dxdy = 3 . 29 zL 2 But evaluating the electric field at z = 0 results in Φ 4 = 0.

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